An artificial cell is made by encapsulating $0.2 \mathrm{~M}$ glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a $0.05 \mathrm{~M}$ solution of $\mathrm{NaCl}$ at $300 \mathrm{~K}$ is ________ $\times 10^{-1}$ bar. (nearest integer).
[Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ]
Assume complete dissociation of $\mathrm{NaCl}$
Explanation:
An artificial cell contains a 0.2 M glucose solution within a semipermeable membrane. When this cell is placed in a 0.05 M NaCl solution at 300 K, we need to determine the osmotic pressure developed, expressed as × 10⁻¹ bar.
Given:
Ideal Gas Constant, R = 0.083 L bar mol⁻¹ K⁻¹
Assume complete dissociation of NaCl.
Explanation
The osmotic pressure ($\pi$) is calculated using the formula:
$ \pi = \left( i_1 C_1 - i_2 C_2 \right) R T $
where:
$i_1$ and $C_1$ are the van 't Hoff factor and concentration for glucose.
$i_2$ and $C_2$ are the van 't Hoff factor and concentration for NaCl.
$R$ is the ideal gas constant.
$T$ is the temperature in Kelvin.
For glucose:
$i_1 = 1$ (glucose does not dissociate)
$C_1 = 0.2$ M
For NaCl:
$i_2 = 2$ (NaCl dissociates into two ions, Na⁺ and Cl⁻)
$C_2 = 0.05$ M
Plugging in the values:
$ \pi = \left(1 \times 0.2 - 2 \times 0.05\right) \times 0.083 \times 300 $
Calculating:
$ \pi = (0.2 - 0.1) \times 0.083 \times 300 $
$ \pi = 0.1 \times 0.083 \times 300 $
$ \pi = 2.5 \text{ bar} $
Hence, the osmotic pressure developed is $25 \times 10^{-1}$ bar.
$2.7 \mathrm{~kg}$ of each of water and acetic acid are mixed. The freezing point of the solution will be $-x^{\circ} \mathrm{C}$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $x=$ ________ (nearest integer)
[Given: Molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$, acetic acid $=60 \mathrm{~g} \mathrm{~mol}^{-1}$
${ }^{\mathrm{K}_{\mathrm{f}}} \mathrm{H}_2 \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$\mathrm{K}_{\mathrm{f}}$ acetic acid: $3.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
freezing point: $\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K}$, acetic acid $=290 \mathrm{~K}$]
Explanation:
$\begin{aligned} & \text { Molality of acetic acid }=\frac{2700}{60} \times \frac{1}{2.7} \mathrm{~mol} / \mathrm{kg} \\ &=16.667 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times 16.667 \\ &=1.86 \times 16.667 \\ &=31 \mathrm{~K} \end{aligned}$
$2.5 \mathrm{~g}$ of a non-volatile, non-electrolyte is dissolved in $100 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$. The solution showed a boiling point elevation by $2^{\circ} \mathrm{C}$. Assuming the solute concentration is negligible with respect to the solvent concentration, the vapor pressure of the resulting aqueous solution is _________ $\mathrm{mm}$ of $\mathrm{Hg}$ (nearest integer)
[Given : Molal boiling point elevation constant of water $\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}$, $1 \mathrm{~atm}$ pressure $=760 \mathrm{~mm}$ of $\mathrm{Hg}$, molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}]$
Explanation:
- Molal boiling point elevation constant of water ($K_b$) = 0.52 K.kg.mol-1
- 1 atm = 760 mm Hg
- Molar mass of water = 18 g.mol-1
First, we calculate the molality (m) of the solution using the boiling point elevation formula:
$ \Delta T_b = K_b \times m $
From the problem, we know:
$ 2 = 0.52 \times m $
Solving for m:
$ m = \frac{2}{0.52} \approx 3.846 \text{ mol/kg} $
Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure:
$ \frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} $
Given that molality (m) is defined as the number of moles of solute per kilogram of solvent:
$ \frac{\Delta P}{P^0} = \frac{m}{1000} \times M_{solvent} $
Substitute the known values:
$ \Delta P = P^0 \times \frac{m}{1000} \times M_{solvent} $
$ = 760 \times \frac{3.846}{1000} \times 18 $
$ = 52.615 \text{ mm Hg} $
Therefore, the vapor pressure of the solution:
$ P_{solution} = P^0 - \Delta P $
$ = 760 - 52.615 $
$ \approx 707.385 \text{ mm Hg} $
Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.
Explanation:
To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\circ} \mathrm{C}$, we can use the freezing point depression equation, which is a colligative property given by:
$ \Delta T_f = i \cdot K_f \cdot m $where:
- $\Delta T_f$ is the depression in the freezing point.
- $i$ is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, $i = 1$.
- $K_f$ is the cryoscopic constant (freezing point depression constant) of water, which is given as $1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}$.
- $m$ is the molality of the solution.
However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:
$ \Delta T_f = -24^{\circ} \mathrm{C} - (0^{\circ} \mathrm{C}) = -24^{\circ} \mathrm{C} $Now, we can solve for $m$:
$ m = \frac{\Delta T_f}{i \cdot K_f} $Substituting the given values:
$ m = \frac{-24^{\circ} \mathrm{C}}{1 \cdot 1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}} $Note that $\mathrm{C}$ and $\mathrm{K}$ are interchangeable when calculating changes in temperature. Now, we can compute the molality:
$ m = \frac{-24}{1.86} $Next, we calculate the molality of the ethylene glycol:
$ m \approx -12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} $Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.
Now we calculate the moles of ethylene glycol required for $18.6 \mathrm{~kg}$ of water:
$ \text{moles of ethylene glycol} = m \cdot \text{mass of water (in kg)} $ $ \text{moles of ethylene glycol} = 12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} \cdot 18.6 \mathrm{~kg} $ $ \text{moles of ethylene glycol} \approx 240.0158 \mathrm{~mol} $Next, we find the mass of ethylene glycol needed using its molar mass:
$ \text{mass of ethylene glycol} = \text{moles of ethylene glycol} \cdot \text{molar mass of ethylene glycol} $Given the molar mass of ethylene glycol is $62 \mathrm{~g} ~\mathrm{mol}^{-1}$, we then convert it to $\mathrm{kg} ~\mathrm{mol}^{-1}$ by dividing by 1000:
$ \text{molar mass of ethylene glycol} = 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} $Now we can calculate the mass:
$ \text{mass of ethylene glycol} = 240.0158 \mathrm{~mol} \cdot 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} $ $ \text{mass of ethylene glycol} \approx 14.88098 \mathrm{~kg} $Therefore, approximately $14.881 \mathrm{~kg}$ of ethylene glycol needs to be added to $18.6 \mathrm{~kg}$ of water to lower the freezing point to $-24^{\circ} \mathrm{C}$.
The osmotic pressure of a dilute solution is $7 \times 10^5 \mathrm{~Pa}$ at $273 \mathrm{~K}$. Osmotic pressure of the same solution at $283 \mathrm{~K}$ is _________ $\times 10^4 \mathrm{Nm}^{-2}$.
Explanation:
To calculate the osmotic pressure of the solution at the new temperature, we can use the formula:
$ \pi = \text{CRT} $
Since the concentration $ C $ and the gas constant $ R $ remain constant, the ratio of the osmotic pressures at two different temperatures is given by:
$ \frac{\pi_1}{\pi_2} = \frac{T_1}{T_2} $
Thus, we can express the osmotic pressure at the second temperature as:
$ \pi_2 = \frac{\pi_1 \cdot T_2}{T_1} $
Substituting the given values:
$ \pi_2 = \frac{7 \times 10^5 \times 283}{273} $
After calculation, this simplifies to:
$ \pi_2 = 72.56 \times 10^4 \, \text{Nm}^{-2} $
The elevation of boiling point for solution in Vessel-1 is ________ $\%$ of the solution in Vessel-2.
Explanation:
To determine the elevation of the boiling point for the solutions in both vessels, we need to use the concept of boiling point elevation. The boiling point elevation is given by the formula:
$ \Delta T_b = i \cdot K_b \cdot m $
Where:
- $\Delta T_b$ is the boiling point elevation
- $i$ is the van't Hoff factor
- $K_b$ is the ebullioscopic constant (which is the same for both solutions since they are dissolved in water)
- $m$ is the molality of the solution
We need to compare the boiling point elevations for both solutions in Vessel-1 and Vessel-2.
For Vessel-1, let the boiling point elevation be $\Delta T_{b1}$, and for Vessel-2, let it be $\Delta T_{b2}$.
Let's denote the molar mass of solute $Y$ as $M_Y$ g/mol. According to the problem, the molar mass of solute $X$ is 80% of that of $Y$, i.e.,
$ M_X = 0.8 \cdot M_Y $
The van't Hoff factor for $X$ is 1.2 times that of $Y$, i.e.,
$ i_X = 1.2 \cdot i_Y $
The molality $m$ of each solution can be calculated using the formula:
$ m = \dfrac{w_2}{M \cdot w_1} $
Thus, the molality of $X$ and $Y$ solutions are:
For solute $X$:
$ m_X = \dfrac{w_2}{M_X \cdot w_1} = \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} $
For solute $Y$:
$ m_Y = \dfrac{w_2}{M_Y \cdot w_1} $
Now, substituting the values into the boiling point elevation formula, we get for Vessel-1:
$ \Delta T_{b1} = i_X \cdot K_b \cdot m_X = (1.2 \cdot i_Y) \cdot K_b \cdot \left( \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} \right) $
For Vessel-2:
$ \Delta T_{b2} = i_Y \cdot K_b \cdot m_Y = i_Y \cdot K_b \cdot \left( \dfrac{w_2}{M_Y \cdot w_1} \right) $
To find the ratio of $\Delta T_{b1}$ to $\Delta T_{b2}$:
$ \dfrac{\Delta T_{b1}}{\Delta T_{b2}} = \dfrac{(1.2 \cdot i_Y) \cdot K_b \cdot \left( \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} \right)}{i_Y \cdot K_b \cdot \left( \dfrac{w_2}{M_Y \cdot w_1} \right)} $
Simplifying the ratio:
$ \dfrac{\Delta T_{b1}}{\Delta T_{b2}} = \dfrac{1.2}{0.8} = 1.5 $
This means the elevation of the boiling point for the solution in Vessel-1 is 150% of that in Vessel-2. Thus, the elevation of boiling point for solution in Vessel-1 is 150% of the elevation of the boiling point for the solution in Vessel-2.
What weight of glucose must be dissolved in $100 \mathrm{~g}$ of water to lower the vapour pressure by $0.20 \mathrm{~mm} ~\mathrm{Hg}$ ?
(Assume dilute solution is being formed)
Given : Vapour pressure of pure water is $54.2 \mathrm{~mm} ~\mathrm{Hg}$ at room temperature. Molar mass of glucose is $180 \mathrm{~g} \mathrm{~mol}^{-1}$
A. The elevation in boiling point temperature of water will be same for $0.1 \mathrm{M} \, \mathrm{NaCl}$ and $0.1 \mathrm{M}$ urea.
B. Azeotropic mixtures boil without change in their composition.
C. Osmosis always takes place from hypertonic to hypotonic solution.
D. The density of $32 \% \, \mathrm{H}_{2} \mathrm{SO}_{4}$ solution having molarity $4.09 ~\mathrm{M}$ is approximately $1.26 \mathrm{~g} \mathrm{~mL}^{-1}$
E. A negatively charged sol is obtained when KI solution is added to silver nitrate solution.
Choose the correct answer from the options given below :
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | van't Hoff factor, i | I. | Cryoscopic constant |
| B. | $\mathrm{k_f}$ | II. | Isotonic solutions |
| C. | Solutions with same osmotic pressure | III. | $\mathrm{\frac{Normal\,molar\,mass}{Abnormal\,molar\,mass}}$ |
| D. | Azeotropes | IV. | Solutions with same composition of vapour above it |
Choose the correct answer from the options given below :
In the depression of freezing point experiment
A. Vapour pressure of the solution is less than that of pure solvent
B. Vapour pressure of the solution is more than that of pure solvent
C. Only solute molecules solidify at the freezing point
D. Only solvent molecules solidify at the freezing point
Choose the most appropriate answer from the options given below :
[Given : The density of $30 \%$ (w/v), aqueous solution of glucose is $1.2 \mathrm{~g} \mathrm{~cm}^{-3}$ and vapour pressure of pure water is $24 \mathrm{~mm}~ \mathrm{Hg}$.]
(Molar mass of glucose is $180 \mathrm{~g} \mathrm{~mol}^{-1}$.)
Explanation:
$\mathrm{Wt.~of~solution} = 100\ \mathrm{mL} \times 1.2\ \mathrm{g/mL} = 120\ \mathrm{g}$
Since the solution is $30 \%$ (w/v), the weight of glucose in $100\ \mathrm{mL}$ of the solution is $30\ \mathrm{g}$. Therefore, the weight of water in $100\ \mathrm{mL}$ of the solution is:
$\mathrm{Wt.~of~water} = 120\ \mathrm{g} - 30\ \mathrm{g} = 90\ \mathrm{g}$
Now, we can use Raoult's law to calculate the vapour pressure of the solution. Raoult's law states that the partial vapour pressure of a solvent in a solution is proportional to its mole fraction in the solution. For a dilute solution, the mole fraction of the solvent can be approximated as the ratio of the number of moles of solvent to the total number of moles in the solution.
Let us assume that we have 1 mole of the solution. The number of moles of glucose in the solution is:
$n_{\mathrm{glucose}} = \frac{30\ \mathrm{g}}{180\ \mathrm{g/mol}} = 0.167\ \mathrm{mol}$
The number of moles of water in the solution is:
$n_{\mathrm{water}} = \frac{90\ \mathrm{g}}{18\ \mathrm{g/mol}} = 5\ \mathrm{mol}$
The total number of moles in the solution is:
$n_{\mathrm{total}} = n_{\mathrm{glucose}} + n_{\mathrm{water}} = 5.167\ \mathrm{mol}$
Let $P$ be the vapour pressure of the solution. According to Raoult's law, we have:
$\frac{P_0 - P}{P} = \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}} = \frac{0.167}{5} = 0.0334$
where $P_0$ is the vapour pressure of pure water, which is given as $24\ \mathrm{mmHg}$.
Simplifying the above equation, we get:
$P = \frac{P_0}{1 + \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}}} = \frac{24\ \mathrm{mmHg}}{1 + \frac{0.167}{5}} = 23.22\ \mathrm{mmHg}$
Therefore, the vapour pressure of the $30\%$ (w/v) aqueous solution of glucose at $25^\circ\mathrm{C}$ is $\boxed{23\ \mathrm{mmHg}}$ (rounded off to the nearest integer).
Sea water contains $29.25 \% ~\mathrm{NaCl}$ and $19 \% ~\mathrm{MgCl}_{2}$ by weight of solution. The normal boiling point of the sea water is _____________ ${ }^{\circ} \mathrm{C}$ (Nearest integer)
Assume $100 \%$ ionization for both $\mathrm{NaCl}$ and $\mathrm{MgCl}_{2}$
Given : $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{NaCl}$ and $\mathrm{MgCl}_{2}$ is 58.5 and 95 $\mathrm{g} \mathrm{~mol}^{-1}$ respectively.
Explanation:
Now, we can calculate the boiling point elevation using the given formula:
ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52
ΔTb = 16.075
The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:
Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C
Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.
Solution of $12 \mathrm{~g}$ of non-electrolyte (A) prepared by dissolving it in $1000 \mathrm{~mL}$ of water exerts the same osmotic pressure as that of $0.05 ~\mathrm{M}$ glucose solution at the same temperature. The empirical formula of $\mathrm{A}$ is $\mathrm{CH}_{2} \mathrm{O}$. The molecular mass of $\mathrm{A}$ is __________ g. (Nearest integer)
Explanation:
1. Osmotic pressure equation:
$\Pi = iMRT$
where $\Pi$ is the osmotic pressure, $i$ is the van't Hoff factor (which is 1 for non-electrolytes), $M$ is the molarity, $R$ is the ideal gas constant (0.0821 L atm/mol K), and $T$ is the temperature in Kelvin.
Since both solutions have the same osmotic pressure at the same temperature, we can set their osmotic pressures equal to each other:
$\Pi_{A} = \Pi_{glucose}$
2. Calculate the osmotic pressure of the 0.05 M glucose solution:
Glucose is a non-electrolyte, so its van't Hoff factor is 1. We don't know the temperature, but since both solutions are at the same temperature, it will cancel out in our calculations.
$\Pi_{glucose} = (1)(0.05 ~\mathrm{M})(R)(T)$
3. Calculate the molarity of compound A:
Since we know that 12 g of compound A is dissolved in 1000 mL of water, we can find the molarity once we know the molecular mass.
Let $x$ be the molecular mass of compound A. Then, the molarity of compound A is:
$M_{A} = \frac{12 ~\mathrm{g}}{x ~\mathrm{g/mol}} \times \frac{1}{1 ~\mathrm{L}} = \frac{12}{x} ~\mathrm{M}$
4. Set the osmotic pressures equal to each other:
$\Pi_{A} = \Pi_{glucose}$
$(1)(\frac{12}{x} ~\mathrm{M})(R)(T) = (1)(0.05 ~\mathrm{M})(R)(T)$
The van't Hoff factors, ideal gas constant, and temperature cancel out:
$\frac{12}{x} = 0.05$
5. Solve for the molecular mass (x) of compound A:
$x = \frac{12}{0.05}$
$x = 240 ~\mathrm{g/mol}$
The molecular mass of compound A is 240 g/mol.
80 mole percent of $\mathrm{MgCl}_{2}$ is dissociated in aqueous solution. The vapour pressure of $1.0 ~\mathrm{molal}$ aqueous solution of $\mathrm{MgCl}_{2}$ at $38^{\circ} \mathrm{C}$ is ____________ $\mathrm{mm} ~\mathrm{Hg.} ~\mathrm{(Nearest} ~\mathrm{integer)}$
Given : Vapour pressure of water at $38^{\circ} \mathrm{C}$ is $50 \mathrm{~mm} ~\mathrm{Hg}$
Explanation:
Hence overall molality will be equal to $=2.6$
$ \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6} $
For dil solution
$ \begin{aligned} & \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}} \\\\ & p=47.66 \simeq 48 \mathrm{~mm} \mathrm{Hg} \end{aligned} $
0.004 M K$_2$SO$_4$ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K$_2$SO$_4$ is ___________ (Nearest integer)
Explanation:
$i \times 0.004 \, \text{M} = 0.01 \, \text{M}$
Here, $i$ is the van't Hoff factor, which accounts for the number of ions the compound dissociates into in the solution. Solving this equation for $i$, we get $i = 2.5$.
For the compound K2SO4, which dissociates into 3 ions (2K+ and 1 SO42-), the formula for $i$ is given by
$i = 1 + (n-1)\alpha$
where $n$ is the number of ions the solute dissociates into and $\alpha$ is the degree of dissociation.
Substituting $n=3$ and $i=2.5$ into this formula and solving for $\alpha$ gives
$\alpha = \frac{i-1}{n-1} = \frac{2.5-1}{3-1} = 0.75$
To convert this into a percentage, we multiply by 100, yielding a percentage dissociation of 75%.
An aqueous solution of volume $300 \mathrm{~cm}^{3}$ contains $0.63 \mathrm{~g}$ of protein. The osmotic pressure of the solution at $300 \mathrm{~K}$ is 1.29 mbar. The molar mass of the protein is ___________ $\mathrm{g} ~\mathrm{mol}^{-1}$
Given : R = 0.083 L bar K$^{-1}$ mol$^{-1}$
Explanation:
The concept we are utilizing to solve this problem is osmotic pressure. Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. For dilute solutions, osmotic pressure behaves similarly to an ideal gas, hence the formula we use is similar to the ideal gas law:
$\Pi = \frac{n}{V}RT$
where:
- $\Pi$ is the osmotic pressure,
- $n$ is the number of moles of solute,
- $V$ is the volume of the solution,
- $R$ is the ideal gas constant, and
- $T$ is the temperature in Kelvin.
First, we convert all given quantities to the appropriate units.
- $\Pi = 1.29$ mbar is equivalent to $1.29 \times 10^{-3}$ bar (since 1 bar = 1000 mbar),
- $V = 300$ cm³ is equivalent to $0.3$ L (since 1 L = 1000 cm³),
- $T = 300$ K,
- $R = 0.083$ L bar K$^{-1}$ mol$^{-1}$.
Instead of the number of moles, we are given the mass of the solute. However, we can replace the moles with mass using the relationship $n = \frac{m}{M}$, where $M$ is the molar mass, and $m$ is the mass of the solute.
So the equation becomes:
$\Pi = \frac{m}{MV}RT$
We are trying to solve for $M$, so rearrange the equation to isolate $M$:
$M = \frac{mRT}{\Pi V}$
Finally, substitute the given values into the equation:
$M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3} \approx 40535 \, \text{g/mol}$
So the molar mass of the protein is approximately 40535 g/mol.
If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be ___________% higher than the expected/theoretical freezing point. (Nearest integer)
Explanation:
The degree of dissociation, often represented as $\alpha$, is the fraction of a mole of a substance that has dissociated into ions in solution. For a weak monobasic acid, this degree of dissociation can increase the number of particles in solution, which can in turn affect colligative properties such as the freezing point.
In this case, a weak monobasic acid, when it dissociates, produces two particles: one $H^{+}$ ion and one anion. So if the degree of dissociation is 0.3 ($\alpha = 0.3$), the average number of particles per molecule of the acid ($i$), also known as the van't Hoff factor, will be $1 + \alpha = 1 + 0.3 = 1.3$.
The decrease in freezing point ($\Delta T_{f}$) is given by the formula $\Delta T_{f} = i \cdot m \cdot K_{f}$, where $m$ is the molality of the solution and $K_{f}$ is the cryoscopic constant of the solvent. Therefore, the observed freezing point depression will be 1.3 times the theoretical freezing point depression for a non-dissociating solute (where $i = 1$).
So, the observed freezing point will be 30% higher than the theoretical freezing point, given the degree of dissociation of 0.3.
If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio $2: 1$ and their enthalpy of vaporizations are in the ratio $1: 2$, then the boiling point elevation constant of $\mathrm{X}$ is $\underline{\mathrm{m}}$ times the boiling point elevation constant of Y. The value of m is ____________ (nearest integer)
Explanation:
$K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}$
where:
- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)
- $T_{b}$ is the boiling point of the solvent (in K)
- $\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol)
According to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2.
Let's denote the boiling point of Y as $T_{b}(Y)$ and the boiling point of X as $T_{b}(X)$, and likewise for the enthalpy of vaporization $\Delta H_{vap}(Y)$ and $\Delta H_{vap}(X)$.
We then have:
$T_{b}(X) = 2T_{b}(Y)$ and $\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)$
We can substitute these values into the equation for $K_{b}$:
$K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))$
and
$K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))$
Comparing these two equations:
$K_{b}(X) / K_{b}(Y) = [4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))] / [R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))] $
This simplifies to:
$K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8$
So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.
The vapour pressure vs. temperature curve for a solution solvent system is shown below.
The boiling point of the solvent is __________ ${ }^{\circ} \mathrm{C}$.
Explanation:
The more volatile liquid evaporates fast as compared to the less volatile liquid at a low temperature because the volume increases with respect to temperature so it has a low boiling point.
Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is / are ___________.
A. $1 ~\mathrm{M}$ aq. $\mathrm{NaCl}$ and $2 ~\mathrm{M}$ aq. urea
B. $1 ~\mathrm{M}$ aq. $\mathrm{CaCl}_{2}$ and $1.5 ~\mathrm{M}$ aq. $\mathrm{KCl}$
C. $1.5 ~\mathrm{M}$ aq. $\mathrm{AlCl}_{3}$ and $2 ~\mathrm{M}$ aq. $\mathrm{Na}_{2} \mathrm{SO}_{4}$
D. $2.5 ~\mathrm{M}$ aq. $\mathrm{KCl}$ and $1 ~\mathrm{M}$ aq. $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$
Explanation:
We say that two solutions are isotonic (at the same temperature) if they have the same osmotic pressure, $\pi$. For dilute solutions at the same temperature $T$,
$ \pi = i\, M\, R\, T, $
where
$M$ = molarity of the solution,
$i$ = van 't Hoff factor (number of particles the solute dissociates into),
$R$ = universal gas constant,
$T$ = absolute temperature.
Since $R$ and $T$ are common for both solutions (same temperature), two solutions will be isotonic if
$ i_1 \, M_1 \;=\; i_2 \, M_2. $
Let us check each pair:
1. Pair A: $1\,\mathrm{M}$ NaCl and $2\,\mathrm{M}$ urea
NaCl dissociates into $ \mathrm{Na}^+ $ and $ \mathrm{Cl}^- $.
Hence $i(\mathrm{NaCl}) = 2$.
So $i \, M = 2 \times 1 = 2.$
Urea ($\mathrm{(NH_2)_2CO}$) does not dissociate in water.
Hence $i(\mathrm{urea}) = 1$.
So $i \, M = 1 \times 2 = 2.$
Both solutions have $i M = 2.$
$\therefore$ They are isotonic.
2. Pair B: $1\,\mathrm{M}$ $\mathrm{CaCl_2}$ and $1.5\,\mathrm{M}$ $\mathrm{KCl}$
Calcium chloride ($\mathrm{CaCl_2}$) dissociates into $ \mathrm{Ca}^{2+} $ and $ 2\,\mathrm{Cl}^- $.
Hence $i(\mathrm{CaCl_2}) = 3.$
So $i \, M = 3 \times 1 = 3.$
Potassium chloride ($\mathrm{KCl}$) dissociates into $ \mathrm{K}^+ $ and $ \mathrm{Cl}^- $.
Hence $i(\mathrm{KCl}) = 2.$
So $i \, M = 2 \times 1.5 = 3.$
Both solutions have $i M = 3.$
$\therefore$ They are isotonic.
3. Pair C: $1.5\,\mathrm{M}$ $\mathrm{AlCl_3}$ and $2\,\mathrm{M}$ $\mathrm{Na_2SO_4}$
Aluminum chloride ($\mathrm{AlCl_3}$) dissociates into $ \mathrm{Al}^{3+} $ and $ 3\,\mathrm{Cl}^- $.
Hence $i(\mathrm{AlCl_3}) = 4.$
So $i \, M = 4 \times 1.5 = 6.$
Sodium sulfate ($\mathrm{Na_2SO_4}$) dissociates into $ 2\,\mathrm{Na}^+ $ and $ \mathrm{SO_4}^{2-} $.
Hence $i(\mathrm{Na_2SO_4}) = 3.$
So $i \, M = 3 \times 2 = 6.$
Both solutions have $i M = 6.$
$\therefore$ They are isotonic.
4. Pair D: $2.5\,\mathrm{M}$ $\mathrm{KCl}$ and $1\,\mathrm{M}$ $\mathrm{Al_2(SO_4)_3}$
Potassium chloride ($\mathrm{KCl}$) has $i(\mathrm{KCl}) = 2.$
So $i \, M = 2 \times 2.5 = 5.$
Aluminum sulfate $\mathrm{Al_2(SO_4)_3}$ dissociates into
$ 2\,\mathrm{Al}^{3+} \;+\; 3\,\mathrm{SO_4}^{2-} \quad \Longrightarrow i(\mathrm{Al_2(SO_4)_3}) = 2 + 3 = 5. $
So $i \, M = 5 \times 1 = 5.$
Both solutions have $i M = 5.$
$\therefore$ They are isotonic.
Conclusion
All four given pairs $(A, B, C, D)$ satisfy $i_1 M_1 = i_2 M_2$. Therefore, all of them are isotonic pairs.
Hence, the number of isotonic pairs is:
$ \boxed{4}. $
Mass of Urea $\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)$ required to be dissolved in $1000 \mathrm{~g}$ of water in order to reduce the vapour pressure of water by $25 \%$ is _________ g. (Nearest integer)
Given: Molar mass of N, C, O and H are $14,12,16$ and $1 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively
Explanation:
Given:
- Vapor pressure reduction: $25\%$ ($0.75$ times the vapor pressure of pure water)
- Molar mass of water ($\text{H}_2\text{O}$): $18 \, \text{g/mol}$
- Mass of solvent (water): $1000 \, \text{g}$
Using Raoult's law:
$\frac{P^0 - P_s}{P_s} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{\frac{x}{M_{\text{urea}}}}{\frac{1000}{M_{\text{water}}}} = \frac{P^0 - 0.75P^0}{0.75P^0}$
Solving for (x):
$\frac{x}{60} = \frac{10000}{9}$
$x = 1111.11111 \approx 1111 \, \text{g}$
So, the mass of urea required to be dissolved in $1000 \, \text{g}$ of water is $1111 \, \text{g}$.
$20 \%$ of acetic acid is dissociated when its $5 \mathrm{~g}$ is added to $500 \mathrm{~mL}$ of water. The depression in freezing point of such water is _________ $\times 10^{-3}{ }^{\circ} \mathrm{C}$.
Atomic mass of $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$ are 12,1 and 16 a.m.u. respectively.
[Given : Molal depression constant and density of water are $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $1 \mathrm{~g} \mathrm{~cm}^{-3}$ respectively.]
Explanation:
25 mL of an aqueous solution of KCl was found to require 20 mL of 1 M $\mathrm{AgNO_3}$ solution when titrated using $\mathrm{K_2CrO_4}$ as an indicator. What is the depression in freezing point of KCl solution of the given concentration? _________ (Nearest integer).
(Given : $\mathrm{K_f=2.0~K~kg~mol^{-1}}$)
Assume 1) 100% ionization and 2) density of the aqueous solution as 1 g mL$^{-1}$
Explanation:
$ \begin{aligned} M & =\frac{20}{25}=\frac{4}{5}=0.8 \\\\ \Delta T_{f} & =(\mathrm{i})\left(\mathrm{K}_{\mathrm{f}}\right)(\mathrm{m}) \\\\ & =(2)(2)\left(\frac{4}{5}\right)=\frac{16}{5}=3.2 \end{aligned} $
At $27^{\circ} \mathrm{C}$, a solution containing $2.5 \mathrm{~g}$ of solute in $250.0 \mathrm{~mL}$ of solution exerts an osmotic pressure of $400 \mathrm{~Pa}$. The molar mass of the solute is ___________ $\mathrm{g} \mathrm{~mol}^{-1}$ (Nearest integer)
(Given : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{~bar} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
$ \begin{aligned} 400 & =\frac{2.5}{\mathrm{m_w}} \times 4 \times\left(.083 \times 10^5\right) \times 300 \\\\ \mathrm{m_w} & =\frac{10 \times 0.083 \times 3}{4} \times 10^5 \\\\ & =62250 \end{aligned} $
Given $\mathrm{K}_{f}=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Explanation:
$\mathrm{\Delta T_f=K_f~i~m}$
$ = 1.8 \times 2.67 \times {{{{38} \over {98}}} \over {0.062}} = 30$
$\therefore$ It freeze at $273-30=243$ $\mathrm{K}$
A solution containing $2 \mathrm{~g}$ of a non-volatile solute in $20 \mathrm{~g}$ of water boils at $373.52 \mathrm{~K}$. The molecular mass of the solute is ___________ $\mathrm{g} ~\mathrm{mol}^{-1}$. (Nearest integer)
Given, water boils at $373 \mathrm{~K}, \mathrm{~K}_{\mathrm{b}}$ for water $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Explanation:
$\mathrm{\Delta T_b=K_b.m}$
(0.52) = (0.52) (m)
$\mathrm{m=1=\frac{2(1000)}{(mw)(20)}}$
mw = 100
Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15$^\circ$C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at $-0.8^\circ$ C. The solubility product of PbCl$_2$ formed is __________ $\times$ 10$^{-6}$ at 298 K. (Nearest integer)
Given : $\mathrm{K_b=0.5}$ K kg mol$^{-1}$ and $\mathrm{K_f=1.8}$ K kg mol$^{-1}$. Assume molality to the equal to molarity in all cases.
Explanation:
$ \begin{aligned} & \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\ & \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \cdot \mathrm{m} \\\\ & 0.8=(0.4+3 s) 1.8 \\\\ & s=0.0148 \\\\ & \therefore \text { solubility product }=4 \mathrm{~s}^{3} \\\\ & =4 \times(0.0148)^{3} \\\\ & \approx 13 \times 10^{-6} \end{aligned} $
The number of pairs of the solutions having the same value of the osmotic pressure from the following is _________.
(Assume 100% ionization)
A. 0.500 $\mathrm{M~C_2H_5OH~(aq)}$ and 0.25 $\mathrm{M~KBr~(aq)}$
B. 0.100 $\mathrm{M~K_4[Fe(CN)_6]~(aq)}$ and 0.100 $\mathrm{M~FeSO_4(NH_4)_2SO_4~(aq)}$
C. 0.05 $\mathrm{M~K_4[Fe(CN)_6]~(aq)}$ and 0.25 $\mathrm{M~NaCl~(aq)}$
D. 0.15 $\mathrm{M~NaCl~(aq)}$ and 0.1 $\mathrm{M~BaCl_2~(aq)}$
E. 0.02 $\mathrm{M~KCl.MgCl_2.6H_2O~(aq)}$ and 0.05 $\mathrm{M~KCl~(aq)}$
Explanation:
The following pairs of solutions have same value of osmotic pressure
(A) $0.500 \mathrm{M} ~\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \mathrm{i}=1$ and $0.25 \mathrm{M} ~\mathrm{KBr}(\mathrm{aq})$ $i=2$
$ \begin{aligned} & \pi_1=0.5 \times 1 \times R T =0.5 R T \\\\ \mathrm{KBr} & \rightleftharpoons \mathrm{K}^{+}+\mathrm{Br}^{-} \\\\ & \pi_2=0.25 \times 2 \times R T=0.5 R T \end{aligned} $
Thus, $\pi_1=\pi_2$
(B) $0.100 \mathrm{M} ~\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathrm{aq}) \mathrm{i}=5$ and $0.100 \mathrm{M}$ $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2}(\mathrm{aq}) \mathrm{i}=5$
$ \begin{array}{ll} \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \quad \pi_1=0.1 \times 5 \times R T=0.5 R T & \\\\ \mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{Fe}^{2+}+2 \mathrm{NH}_4^{+}+2 \mathrm{SO}_4^{2-} & (i=5) \\\\ \pi_2=0.1 \times 5 \times R T=0.5 R T & \end{array} $
Thus, $\pi_1=\pi_2$
C. 0.05 $\mathrm{M~K_4[Fe(CN)_6]~(aq)}$ and 0.25 $\mathrm{M~NaCl~(aq)}$
$ \begin{gathered} {K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 {~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \pi_1=0.05 \times 5 \times R T=0.25 R T \\\\ {NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \quad(i=2) \\\\ \pi_2=0.25 \times 2 R T=0.5 R T \end{gathered} $
Thus, $\pi_1 \neq \pi_2$
(D) $0.15 \mathrm{M}~ \mathrm{NaCl}(\mathrm{aq}) \mathrm{i}=2$ and $0.10 \mathrm{M}~ \mathrm{BaCl}_{2}$ (aq) $i=3$
$ \pi_1=0.15 \times 2 \times R T=0.3 R T$
$ \begin{aligned} \mathrm{BaCl}_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}(i=3) \\\\ \pi_2=0.1 \times 3 \times R T=0.3 R T \end{aligned} $
Thus, $\pi_1=\pi_2$
(E) $0.02 \mathrm{M}~ \mathrm{KCl} . \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \mathrm{i}=5$ and $0.05 \mathrm{M}$ $\mathrm{KCl}(\mathrm{aq}) \mathrm{i}=2$
$ \mathrm{KCl} \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}^{+}+3 \mathrm{Cl}^{-}+\mathrm{Mg}^{2+} \quad(i=5)$
$ \begin{aligned} & \pi_1=5 \times 0.02 R T=0.1 R T \\\\ & 0.05 \mathrm{M} \mathrm{KCl}, \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+\mathrm{Cl}^{-}(i=2) \\\\ & \pi_2=0.05 \times 2 R T=0.1 R T \end{aligned} $
Thus, $\pi_1=\pi_2$
The osmotic pressure of solutions of PVC in cyclohexanone at 300 K are plotted on the graph.
The molar mass of PVC is ____________ g mol$^{-1}$ (Nearest integer)
(Given : R = 0.083 L atm K$^{-1}$ mol$^{-1}$)
Explanation:
If we assume graph between $\frac{\pi}{\mathrm{C}}$ and $\mathrm{C}$, then right graph should be like this in the question,
$ \therefore $ Question's graph is wrong.
Assuming $\pi $ vs C graph,
$\begin{aligned} & \pi=C R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times \frac{\mathrm{M}}{\mathrm{M}} \times \mathrm{RT} \\\\ & \Rightarrow \pi=\frac{\text { mass }}{\text { volume }} \times \frac{\mathrm{RT}}{\mathrm{M}} \\\\ & \Rightarrow \pi(\mathrm{atm})=\frac{R T}{M} \times C\left(\mathrm{g} \mathrm{lit}^{-1}\right)\end{aligned}$
$\begin{aligned} & \text {So, Slope }=\frac{\mathrm{RT}}{\mathrm{M}}=\frac{0.083 \times 300}{\mathrm{M}}=6 \times 10^{-4} \\\\ & \therefore \mathrm{M}=\frac{0.083 \times 300}{6 \times 10^{-4}}=\frac{830 \times 300}{6} \\\\ & =41,500 \mathrm{g} / \mathrm{mole}\end{aligned}$
The total pressure observed by mixing two liquids A and B is 350 mm Hg when their mole fractions are 0.7 and 0.3 respectively.
The total pressure becomes 410 mm Hg if the mole fractions are changed to 0.2 and 0.8 respectively for A and B. The vapour pressure of pure A is __________ mm Hg. (Nearest integer)
Consider the liquids and solutions behave ideally.
Explanation:
Let $V . P$ of pure $B$ be $P_B^0$
When $\mathrm{X}_{\mathrm{A}}=0.7 \& \mathrm{X}_{\mathrm{B}}=0.3$
$ \begin{aligned} & \mathrm{P}_{\mathrm{s}}=350 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.7+\mathrm{P}_{\mathrm{B}}^0 \times 0.3=350 \end{aligned} $
When $\mathrm{X}_{\mathrm{A}}=0.2 \& \mathrm{X}_{\mathrm{B}}=0.8$
$ \begin{aligned} & P_{\mathrm{s}}=410 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.2+\mathrm{P}_{\mathrm{B}}^0 \times 0.8=410 \end{aligned} $
Solving (i) and (ii)
$ \begin{aligned} & \mathrm{P}_{\mathrm{A}}^0=314 \mathrm{~mm} \mathrm{Hg} \\\\ & \mathrm{P}_{\mathrm{B}}^0=434 \mathrm{~mm} \mathrm{Hg} \\\\ & \therefore \text{Answer} = 314 \end{aligned} $
[Use: Molar mass of urea $=60 \mathrm{~g} \mathrm{~mol}^{-1}$; gas constant, $\mathrm{R}=62$ L Torr $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$;
Assume, $\Delta_{\text {mix }} \mathrm{H}=0, \Delta_{\text {mix }} \mathrm{V}=0$ ]
Explanation:
Find the weight of urea in the 0.2 molal solution :
Given a 0.2 molal solution, there are 0.2 moles of urea in 1000 g of solvent. The weight of urea is thus $0.2 \, \text{mol} \times 60 \, \text{g/mol} = 12 \, \text{g}$.Find the weight of the solution :
The total weight of the solution is the weight of the solvent plus the weight of the solute, or $1000 \, \text{g} + 12 \, \text{g} = 1012 \, \text{g}$.Find the volume of the solution :
Given the density of the solution, we can find its volume by dividing the total weight of the solution by the density: $\frac{1012 \, \text{g}}{1.012 \, \text{g/mL}} = 1000 \, \text{mL}$.Find the amount of urea in 50 mL of the 0.2 molal solution :
If 1000 mL of the solution contains 0.2 moles of urea, then 50 mL of the solution contains $\frac{0.2 \, \text{mol} \times 50 \, \text{mL}}{1000 \, \text{mL}} = 0.01 \, \text{mol}$ of urea.Find the amount of urea in the 250 mL solution :
The 250 mL solution contains $0.06 \, \text{g}$ of urea, or $\frac{0.06 \, \text{g}}{60 \, \text{g/mol}} = 0.001 \, \text{mol}$.Find the total concentration of the solution :
After mixing, the total volume of the solution is $50 \, \text{mL} + 250 \, \text{mL} = 300 \, \text{mL}$, and the total amount of urea is $0.01 \, \text{mol} + 0.001 \, \text{mol} = 0.011 \, \text{mol}$.
So, the concentration of the solution is $\frac{0.011 \, \text{mol}}{300 \, \text{mL}} \times 1000 \, \text{mL/L} = 0.0366 \, \text{M}$.Find the osmotic pressure of the solution :
Finally, the osmotic pressure $\pi$ of the solution can be found using the formula $\pi = CRT$, where $C$ is the concentration, $R$ is the gas constant, and $T$ is the temperature. Substituting the given and calculated values,
we have $\pi = 0.0366 \, \text{M} \times 62 \, \text{L Torr K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} = 682 \, \text{Torr}$.
Boiling point of a $2 \%$ aqueous solution of a non-volatile solute A is equal to the boiling point of $8 \%$ aqueous solution of a non-volatile solute B. The relation between molecular weights of A and B is
Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y, respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1 : 4. The ratio of molar masses of X and Y is
The depression in freezing point observed for a formic acid solution of concentration $0.5 \mathrm{~mL} \mathrm{~L}^{-1}$ is $0.0405^{\circ} \mathrm{C}$. Density of formic acid is $1.05 \mathrm{~g} \mathrm{~mL}^{-1}$. The Van't Hoff factor of the formic acid solution is nearly : (Given for water $\mathrm{k}_{\mathrm{f}}=1.86\, \mathrm{k} \,\mathrm{kg}\,\mathrm{mol}^{-1}$ )
For a solution of the gases A, B, C and D in water at 298 K, the values of Henry's law constant (KH) are 30.40, 2.34, 1.56 $\times$ 10$-$5 and 0.513 k bar respectively. In the given graph, the lines marked as 'p' and 's' correspond respectively to :
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : At 10$^\circ$C, the density of a 5 M solution of KCl [atomic masses of K & Cl are 39 & 35.5 g mol$-$1 respectively], is 'x' g ml$-$1. The solution is cooled to $-$21$^\circ$C. The molality of the solution will remain unchanged.
Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature.
In the light of the above statements, choose the correct answer from the options given below :
Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water, it depresses the freezing point by 0.2$^\circ$C. The percentage association of solute A in water, is :
[Given : Molar mass of A = 93 g mol$-$1. Molal depression constant of water is 1.86 K kg mol$-$1.]
$1.80 \mathrm{~g}$ of solute A was dissolved in $62.5 \mathrm{~cm}^{3}$ of ethanol and freezing point of the solution was found to be $155.1 \mathrm{~K}$. The molar mass of solute A is ________ g $\mathrm{mol}^{-1}$.
[Given : Freezing point of ethanol is 156.0 K.
Density of ethanol is 0.80 g cm$-$3.
Freezing point depression constant of ethanol is 2.00 K kg mol$-$1]
Explanation:
$\Delta T_{f}=k_{f} \times m$
$ \begin{aligned} &0.9=2\left[\frac{1.8 \times 1000}{M_{\text {Solute }} \times 50}\right] \\ &M_{\text {Solute }}=\left(\frac{2 \times 1.8 \times 1000}{0.9 \times 50}\right)=80 \end{aligned} $
If $\mathrm{O}_{2}$ gas is bubbled through water at $303 \mathrm{~K}$, the number of millimoles of $\mathrm{O}_{2}$ gas that dissolve in 1 litre of water is __________. (Nearest Integer)
(Given : Henry's Law constant for $\mathrm{O}_{2}$ at $303 \mathrm{~K}$ is $46.82 \,\mathrm{k}$ bar and partial pressure of $\mathrm{O}_{2}=0.920$ bar)
(Assume solubility of $\mathrm{O}_{2}$ in water is too small, nearly negligible)
Explanation:
$0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}$
$0.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18}$
$0.920=46.82 \times n_{O_{2}}$
$\mathrm{P}=\frac{0.920}{46.82 \times 18}=n_{O_{2}}$
$\Rightarrow 1.09 \times 10^{-3} \,n_{O_{2}}$
$\Rightarrow \mathrm{m}\, \mathrm{mol}\, \mathrm{of}\, \mathrm{O}_{2}=1$
A gaseous mixture of two substances A and B, under a total pressure of $0.8$ atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is $0.5$ in the vapour phase and $0.2$ in the liquid phase. The vapour pressure of pure liquid $\mathrm{A}$ is __________ atm. (Nearest integer)
Explanation:
We know that $P_{A}=Y_{A} \times P_{T}$
$ P_{A}=0.5 \times 0.8=0.4 $
Now $P_{A}=X_{A} \times P_{A}^{\circ} \Rightarrow P_{A}^{\circ}=\frac{0.4}{0.2}=2 \mathrm{~atm}$
$150 \mathrm{~g}$ of acetic acid was contaminated with $10.2 \mathrm{~g}$ ascorbic acid $\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)$ to lower down its freezing point by $\left(x \times 10^{-1}\right)^{\circ} \mathrm{C}$. The value of $x$ is ___________. (Nearest integer)
[Given $\mathrm{K}_{f}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; molar mass of ascorbic acid $=176 \mathrm{~g} \mathrm{~mol}^{-1}$]
Explanation:
M.wt. of Ascorbic acid $=176 \mathrm{~g}$
$ \Delta T_{f}=K_{f} m $
$ \begin{aligned} \Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\ \Delta T_{f} &=1.506 \\\\ &=15.06 \times 10^{-1} \\\\ &=15 \end{aligned} $
When a certain amount of solid A is dissolved in $100 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$ to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is $23.76 \,\mathrm{mmHg}$. The number of moles of solute A added is _____________. (Nearest Integer)
Explanation:
$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0} \sim \frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $
$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $
${ }^{\mathrm{n}}$ solute $\sim \frac{{ }^{\mathrm{n}} \text { solvent }}{2}=\frac{100}{18 \times 2}=2.78 \mathrm{~mol}$
More accurate approach:
$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $
$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0 / 2}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^n \text { solvent }} $
${ }^n$ solute $={ }^n$ solvent $=\frac{100}{18}=5.55 \mathrm{~mol}$
The elevation in boiling point for 1 molal solution of non-volatile solute A is $3 \mathrm{~K}$. The depression in freezing point for 2 molal solution of $\mathrm{A}$ in the same solvent is 6 $K$. The ratio of $K_{b}$ and $K_{f}$ i.e., $K_{b} / K_{f}$ is $1: X$. The value of $X$ is [nearest integer]
Explanation:
Elevation in boiling point is given by
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$
$ 3=\mathrm{K}_{\mathrm{b}} \times 1 $ ... (1)
Molality of $(A)$ in the same solvent $=2$
Depression in freezing point is given by
$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \\ &6=\mathrm{K}_{\mathrm{f}} \times 2 \quad\quad\quad{...(2)} \\ &\text { Dividing (1) by (2) } \\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=\frac{1}{\mathrm{X}}=\frac{1}{1} \\ &\therefore \quad \mathrm{X}=1 \end{aligned} $
Elevation in boiling point for 1.5 molal solution of glucose in water is 4 K. The depression in freezing point for 4.5 molal solution of glucose in water is 4 K. The ratio of molal elevation constant to molal depression constant (Kb/Kf) is _________.
Explanation:
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198$^\circ$C. The percentage of dissociation of the acid is ___________. (Nearest integer)
[Given : Density of acetic acid is 1.02 g mL$-$1, Molar mass of acetic acid is 60 g mol$-$1, Kf(H2O) = 1.85 K kg mol$-$1]
Explanation:
Moles of solute $($ acetic acid $)=\frac{1.2 \times 1.02}{60}$
As moles of solute are very less.
So, take molarity and molality the same.
$0.0198=\mathrm{i} \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}$
$\mathrm{i}=1.05$
$\alpha=\frac{i-1}{n-1}=\frac{0.05}{1}= 0.05 = 5\text{%}$
2.5 g of protein containing only glycine (C2H5NO2) is dissolved in water to make 500 mL of solution. The osmotic pressure of this solution at 300 K is found to be 5.03 $\times$ 10$-$3 bar. The total number of glycine units present in the protein is ____________.
(Given : R = 0.083 L bar K$-$1 mol$-$1)
Explanation:
$ \pi=\mathrm{icR} \mathrm{T} $
$5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300$
Molar mass of protein $=24751.5 \mathrm{~g} / \mathrm{mol}$
Number of glycine units in protein $=\frac{24751.5}{75}$
$ =330 $
The vapour pressures of two volatile liquids A and B at 25$^\circ$C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is ${x \over {17}}$. The value of x is ______________.
Explanation:
A solution containing 2.5 $\times$ 10$-$3 kg of a solute dissolved in 75 $\times$ 10$-$3 kg of water boils at 373.535 K. The molar mass of the solute is ____________ g mol$-$1. [nearest integer] (Given : Kb(H2O) = 0.52 K kg mol$-$1 and boiling point of water = 373.15 K)
Explanation:
$ \begin{aligned} &\mathrm{W}_{\text {solvent }}=75 \times 10^{-3} \mathrm{~kg} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =373.535-373.15 \\\\ &=0.385 \mathrm{~K} \\\\ &\mathrm{~K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right) =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =\frac{\mathrm{K}_{\mathrm{b}} \times 10^{3} \times \mathrm{W}_{\text {solute }}}{\mathrm{M}_{\text {solute }} \times \mathrm{W}_{\text {solvent }}} \\\\ &\mathrm{M}_{\text {solute }} =\frac{0.52 \times 10^{3} \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385} \\\\ &=45.02 \\\\ & \approx 45 \end{aligned} $
2 g of a non-volatile non-electrolyte solute is dissolved in 200 g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 8. The elevation in boiling points of A and B are in the ratio ${x \over y}$ (x : y). The value of y is ______________. (Nearest integer)
Explanation:
$ \begin{aligned} &\frac{\left(\Delta T_{b}\right)_{A}}{\left(\Delta T_{b}\right)_{B}}=\frac{\left(k_{b}\right)_{A}}{\left(k_{b}\right)_{B}} \\\\ &=\frac{1}{8}=\frac{x}{y} \\\\ &\therefore y=8 \end{aligned} $
The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol$-$1 in 200 mL of water at 27$^\circ$C is ______________ Pa. [integer value]
(use R = 0.083 L bar mol$-$1 K$-$1)
Explanation:
$\pi=\frac{2 \times 1000}{60 \times 10^{3} \times 200} \times .083 \times 300$
$\pi=0.00415 \mathrm{~atm}$
$\pi=415 \mathrm{~Pa}$