Solutions

Density of the solution:

$ \text{Density} = 1.00 \, \text{g cm}^{-3} = 1000 \, \text{kg m}^{-3} $

Height of the solution column ($h$):

$ h = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} $

Acceleration due to gravity ($g$):

$ g = 10 \, \text{m s}^{-2} $

Calculating osmotic pressure ($\pi$):

Osmotic pressure is given by the equation:

$ \pi = h \cdot \rho \cdot g $

Substituting the known values:

$ \pi = 2 \times 10^{-2} \times 1000 \times 10 = 200 \, \text{N m}^{-2} $

Relating osmotic pressure to concentration and molar mass:

The equation for osmotic pressure in terms of concentration and molar mass is:

$ \pi = cRT $

where $c = \frac{2000}{M}$ as the concentration in g/dm$^3$ needs to be converted to mol/dm$^3$ by dividing by the molar mass $M$. Incorporating the given temperature and universal gas constant, we have:

$ 200 = \left(\frac{2000}{M}\right) \cdot 8.3 \cdot 300 $

Solving for the molar mass $M$:

Rearrange and solve the equation to find $M$:

$ M = 24900 \, \text{g mol}^{-1} = 2.49 \times 10^4 \, \text{g mol}^{-1} $

Thus, the value of $X$ is $2.49$.

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.

1. 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 $0.2 \, \text{mol} \times 60 \, \text{g/mol} = 12 \, \text{g}$.




2. 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 $1000 \, \text{g} + 12 \, \text{g} = 1012 \, \text{g}$.




3. 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: $\frac{1012 \, \text{g}}{1.012 \, \text{g/mL}} = 1000 \, \text{mL}$.




4. 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.




5. Find the amount of urea in the 250 mL solution :
   
   The 250 mL solution contains $0.06 \, \text{g}$ of urea, or $\frac{0.06 \, \text{g}}{60 \, \text{g/mol}} = 0.001 \, \text{mol}$.




6. 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 $\frac{0.011 \, \text{mol}}{300 \, \text{mL}} \times 1000 \, \text{mL/L} = 0.0366 \, \text{M}$.




7. 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 $\pi = 0.0366 \, \text{M} \times 62 \, \text{L Torr K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} = 682 \, \text{Torr}$.

Vapour pressure of solution $\left(\mathrm{P}_{\mathrm{A}}\right)$ $ =59.724 \mathrm{~mm} \text { of } \mathrm{Hg} $

Vapour pressure of pure water $\left(\mathrm{P}_{\mathrm{A}}^{\circ}\right)$ $ =60.000 \mathrm{~mm} \text { of } \mathrm{Hg} $

Also, $0.1 \mathrm{~mol}$ of an ionic solid is dissolved in $1.8 \mathrm{~kg}$ of water and salt remains $90 \%$ dissocated in the solution.



So, total number of moles $=0.01+0.09 a$ of non-volatile particles.

Now, mass of water $=1.8 \mathrm{~kg}=1.8 \times 1.8 \times 1000 \mathrm{~g}$

Molar mass of water $=18 \mathrm{~g}$

Moles of water $=\frac{1.8 \times 1000}{18}=100$ moles

Using the colligative property, relative lowering in vapour pressure,

$ \frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=x_{\mathrm{A}} $

$ \Rightarrow $ $\frac{60-59.724}{60} =\frac{0.01+0.09 a}{100} $

$ \Rightarrow $ $ \frac{0.276}{60} =\frac{0.01+0.09 a}{100} $

$ \Rightarrow $ $ \frac{27.6}{60} =0.01+0.09 a $

$ \Rightarrow $ $ 0.46 =0.01+0.09 a $

$ \Rightarrow $ $ 0.09 a =0.45 $

$ \Rightarrow $ $ a =\frac{0.45}{0.89} $

$ \Rightarrow $ $ a =5 $

So, the number of ions present per formula unit of the ionic salt is 5 .

$ \begin{aligned} & \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \underset{0.1 ~ m}{\mathrm{Ag}^{+}(\mathrm{aq})}+\underset{0.1 ~m}{\mathrm{NO}_3^{-}(\mathrm{aq})}\\\\ & \Delta \mathrm{T}_{\mathrm{b}}=0.2 \times 0.5 \\\\ & =0.1^{\circ} \mathrm{C}=0.1 \mathrm{~K} \end{aligned} $

Boiling point of solution $=100.1^{\circ} \mathrm{C} =\mathrm{X} $

$ \begin{aligned} & \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{NO}_3^{-}(\mathrm{aq}) \\\\ & 0.05 \mathrm{~m} \quad\quad\quad\quad\quad 0.05 \mathrm{~m} \quad\quad 0.05 \mathrm{~m} \\\\ & \mathrm{BaCl}_2(\mathrm{aq}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \\\\ & 0.05 \mathrm{~m} \quad \quad\quad\quad\quad0.05 \mathrm{~m} \quad\quad 0.1 \mathrm{~m} \end{aligned} $

$\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$combine to form $\mathrm{AgCl}$ precipitate

$ \begin{array}{cccc} & \mathrm{Ag}^{+}(\mathrm{aq})+ & \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow & \mathrm{AgCl}(\mathrm{s}) \\\\ \mathrm{t}=0 & 0.05 \mathrm{~m} & 0.1 \mathrm{~m} \\\\ \mathrm{t}=\infty & 0 & 0.05 \mathrm{~m} \end{array} $

In final solution total concentration of all ions :

$ \begin{aligned} & {\left[\mathrm{Cl}^{-}\right]+\left[\mathrm{NO}_3^{-}\right]+\left[\mathrm{Ba}^{2+}\right]=0.05+0.05+0.05} \\\\ & \begin{aligned} \Delta \mathrm{T}_{\mathrm{b}} & =0.5 \times 0.15 \\\\ & =0.15 \mathrm{~m} \\\\ & =0.075^{\circ} \mathrm{C} \end{aligned} \end{aligned} $

B.P. of solution ' $\mathrm{B}$ ' $=100.075^{\circ} \mathrm{C}$

B.P. of solution ' $\mathrm{A}^{\prime}=100.1^{\circ} \mathrm{C}$

$|y|=100.1-100.075$

$ =0.025=2.5 \times 10^{-2} $

Using Raoult's law equation for a mixture of volatile liquids.

${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$

$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$ ....(i)

$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$ ....(ii)

By solving equation (i) and (ii)

$p_A^o$ = 0.6 bar and $p_B^o$ = 0.2 bar

Thus, the vapour pressure of pure liquid B in bar is 0.2.

When 0.5 g of non-volatile solute dissolve into 39 gm of benzene then relative lowering of vapour pressure occurs. Hence, vapour pressure decreases from 650 mmHg to 640 mmHg.

Given, vapour pressure of solvent (p$^\circ $) = 650 mmHg

Vapour pressure of solution (p_s) = 640 mmHg

Weight of non-volatile solute = 0.5 g

Weight of solvent (benzene) = 39 g

From relative lowering of vapour pressure,

${{p^\circ - {p_s}} \over {p^\circ }} = {x_{Solute}} = {{{n_{solute}}} \over {{n_{solute}} + {n_{solvent}}}}$

${{650 - 640} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + {{39} \over {78}}}}$

${{10} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + 0.5}}$

$0.5 + 0.5 \times molar\,mass = 65 \times 0.5$

$ \therefore $ Molar mass of solute = 64 g

From molal depression of freezing point,

$\Delta {T_f} = {K_f} \times molality$

$ = {{{K_f} \times {w_{solute}}} \over {{{(MW)}_{solute}} \times {w_{solvent}}}}$

$ \Rightarrow $ $ \Delta {T_f} = 5.12 \times {{0.5 \times 1000} \over {64 \times 39}} $

$\Rightarrow \Delta {T_f} = 1.02K$

We know,

${p_{Total}} = p_A^o \times {\chi _A} + p_B^o \times {\chi _B}$

and for equimolar solutions ${\chi _A} = {\chi _B} = 1/2$

Given, ${p_{total}} = 45$ torr for equimolar solution and $p_A^o = 20$ torr

So, $45 = p_A^o \times {1 \over 2} + p_B^o \times {1 \over 2} = {1 \over 2}(p_A^o + p_B^o)$

or $p_A^o + p_B^o = 90$ torr ....... (i)

But we know $p_A^o = 20$ torr

So, $p_B^o = 90 - 20 = 70$ torr (From Eq. (i))

Now, for the new solution from the same formula

${p_{total}} = 22.5$ torr

$22.5 = 20{\chi _A} + 70(1 - {\chi _A})$ (As ${\chi _A} + {\chi _B} = 1$)

or $22.5 = 70 - 50{\chi _A}$

So, ${\chi _A} = {{70 - 22.5} \over {50}} = 0.95$

Thus, ${\chi _B} = 1 - 0.95 = 0.05$ (as $\chi$_A + $\chi$_B = 1)

Hence, the ratio

${{{\chi _A}} \over {{\chi _B}}} = {{0.95} \over {0.05}} = 19$

For solvent X, $\Delta$T_bX = K_bX m $\Rightarrow$ 2 = 2K_bX m ..... (1)

For solvent Y, $\Delta$T_bY = K_bY m $\Rightarrow$ 1 = 2K_bY m ..... (2)

Dividing Eq. (1) by Eq. (2), we get

${{{K_{bX}}} \over {{K_{bY}}}} = 2$

After adding solute S, molality is the same for both solutions.

Dimerisation is a association property. And for dimerisation van't Hoff factor

i = 1 + $\left( {{1 \over n} - 1} \right)\alpha $

And for dimerisation n = 2

$ \therefore $ For solvent X, van't Hoff factor $i = 1 - {\alpha \over 2}$

$\Delta {T_{bX}} = i{K_{bX}}m = \left( {1 - {\alpha \over 2}} \right){K_{bX}}m$ ....... (3)

For solvent Y, van't Hoff factor $i = 1 - {{0.7} \over 2} = {{1.3} \over 2}$

$\Delta {T_{bY}} = i{K_{bY}}m \Rightarrow \left( {{{1.3} \over 2}} \right){K_{bY}}m$ ....... (4)

Given that $\Delta {T_{bX}} = 3\Delta {T_{bY}}$. Dividing Eq. (3)/(4), we have

${{\Delta {T_{bX}}} \over {\Delta {T_{bY}}}} = {{\left( {1 - {\alpha \over 2}} \right){K_{bX}}m} \over {\left( {{{1.3} \over 2}} \right){K_{bY}}m}} \Rightarrow {1 \over 3} = {{2(2 - \alpha )} \over {1.3}} \Rightarrow \alpha = 0.05$

It is given that the mole fraction of the solute $\left(x_{\text {solute }}\right)$ is 0.1 , therefore mole fraction of the solvent $\left(x_{\text {solvent }}\right)$ is 0.9 . So we have

$ x_{\text {solute }}=\frac{n_{\text {solute }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.1 $ ......(1)

$ x_{\text {solvent }}=\frac{n_{\text {solvent }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.9 $ ......(2)

where $n_{\text {solute }}$ and $n_{\text {solvent }}$ are the number of moles of solute and solvent, respectively. Dividing Eq. (1) by Eq. (2) gives

$ \frac{x_{\text {solute }}}{x_{\text {solvent }}}=\frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{0.1}{0.9} $ .......(3)

Given that the density of solution is $2 \mathrm{~g} \mathrm{~cm}^{-3}$, we have

$ \begin{aligned} & W_{\text {solution }} =\text { density } \times V_{\text {solution }}=2 \times V_{\text {solution }} \\\\ & \therefore W_{\text {solute }}+W_{\text {solvent }} =2 \times V_{\text {solution }} .......(4) \end{aligned} $

We know that molality is given by

$ m=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times W_{\text {solvent }}} $

Substituting from Eq. (3), we have

$ m=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{W_{\text {solvent }}} $ ......(5)

Molarity is given by

$ M=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times V_{\text {solution }}} $

Substituting from Eq. (3), we have

$ M=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{V_{\text {solution }}} $ ..........(6)

Given that, Molarity $(M)=$ Molality $(m)$;

Therefore, from Eqs. (5) and (6), we get

$ \begin{aligned} & \frac{0.1 \times 1000}{0.9 \times W_{\text {solvent }}} =\frac{0.1 \times 1000}{0.9 \times V_{\text {solution }}} \\\\ & W_{\text {solvent }} =V_{\text {solution }} \end{aligned} $

From Eq. (4), we have

$ \begin{aligned} W_{\text {solvent }} & =\frac{W_{\text {solute }}+W_{\text {solvent }}}{2} \\\\ 2 W_{\text {solvent }} & =W_{\text {solute }}+W_{\text {solvent }} \\\\ W_{\text {solvent }} & =W_{\text {solute }} .........(7) \end{aligned} $

The molecular weight (MW) of the solute can be calculated by dividing the weight of the solute by the number of moles of solute. This can be written as :

$ MW_{\text{solute}} = \frac{W_{\text{solute}}}{n_{\text{solute}}} $

2. Similarly, the molecular weight (MW) of the solvent can be calculated by dividing the weight of the solvent by the number of moles of solvent. This can be written as :

$ MW_{\text{solvent}} = \frac{W_{\text{solvent}}}{n_{\text{solvent}}} $

In these formulas, $MW_{\text{solute}}$ and $MW_{\text{solvent}}$ represent the molecular weights of the solute and solvent respectively, $W_{\text{solute}}$ and $W_{\text{solvent}}$ represent their weights, and $n_{\text{solute}}$ and $n_{\text{solvent}}$ represent the number of moles of solute and solvent respectively.

From Eq. (3), we have

$ \frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{W_{\text {solute }} / M W_{\text {solute }}}{W_{\text {solvent }} / M W_{\text {solvent }}}=\frac{0.1}{0.9} $

Using Eq. (7), we get

$ \frac{M W_{\text {solute }}}{M W_{\text {solvent }}}=9 $

The depression in freezing point is given by

$\Delta$T_f = K_f $\times$ m $\times$ i

0.0558 = 1.86 $\times$ 0.01 $\times$ i

i = 3

Therefore, one mole of complex gives three moles of ions in solution.

Hence, the complex is [Co(NH₃)₅Cl]Cl₂ and the number of Cl^$-$ ions inside the coordination sphere is 1.

The depression of freezing point is an important colligative property, which is influenced by the number of solute particles in a solution. When a solute dissociates or ionizes in a solution, it increases the number of particles in the solution, thereby affecting colligative properties such as the depression of freezing point. The degree of dissociation ($\alpha$) gives us a measure of the extent to which a compound dissociates into its ions. For our case, MX₂ dissociates into one M²⁺ ion and two X^- ions.

Let's initially consider 1 mole of MX₂ is present. Since the degree of dissociation ($\alpha$) is 0.5, this means half of the MX₂ dissociates into its ions, and half remains undissociated.

For the dissociation reaction:

$\text{MX}_{2} \rightarrow \text{M}^{2+} + 2\text{X}^{-}$

If the initial amount of MX₂ is 1 mole, then:

• The amount of MX₂ that dissociates = $\alpha = 0.5$ moles
• The amount of M²⁺ formed = $\alpha = 0.5$ moles (since for every mole of MX₂ that dissociates, 1 mole of M²⁺ is formed)
• The amount of X^- formed = $2\alpha = 2 \times 0.5 = 1$ mole (since for every mole of MX₂ that dissociates, 2 moles of X^- are formed)

The Van't Hoff factor (i) quantifies the effect of solute particles on the colligative properties of a solution. It is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved in the solvent.

$i = \frac{\text{Total number of particles after dissociation}}{\text{Number of moles of solute originally dissolved}}$

Considering the dissociation and the amounts calculated:

• Total number of particles after dissociation = (undissociated MX₂) + (M²⁺) + (X^-) = $(1 - \alpha) + \alpha + 2\alpha$
• Putting the value of $\alpha = 0.5$, we get:

  $i = \frac{(1 - 0.5) + 0.5 + 2(0.5)}{1} = \frac{1 + 1}{1} = 2$

Now, the depression in freezing point ($\Delta T_f$) is directly proportional to the molal concentration of the solute particles and Van't Hoff factor (i):

$\Delta T_f = i \cdot K_f \cdot m$

Where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.

Without ionic dissociation, the value of $i$ would have been 1 (since the solute would not have dissociated into multiple particles). However, due to the dissociation, the value of $i$ has increased to 2. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation, therefore, would be directly equal to the ratio of the Van't Hoff factors:

$\frac{\text{Observed } \Delta T_f}{\text{In the absence of ionic dissociation}} = \frac{2 \cdot K_f \cdot m}{1 \cdot K_f \cdot m} = \frac{2}{1} = 2$

This ratio shows that due to the ionic dissociation of MX₂ into M²⁺ and X^- ions with a degree of dissociation ($\alpha$) of 0.5, the observed depression in freezing point is twice the value it would have been in the absence of ionic dissociation.

Step 1: Calculate moles of phenol.
The number of moles of phenol is ${{75.2} \over {94}} = 0.8$ mol.
(Here, the molar mass of phenol is 94 g/mol, and 75.2 g is dissolved.)

Step 2: Find the molality of the solution.
Molality (m) = ${{0.8} \over {1}} = 0.8$ mol/kg.
(Because 0.8 mol phenol is dissolved in 1 kg of solvent.)

Step 3: Use the formula for depression in freezing point.
We know $\Delta T_f = k_f \times m$
Given: $k_f = 14$, so substitute the values.

$\Delta T_f = 14 \times 0.8 = 11.2~\mathrm{K}$
This is the calculated value if no dimerization happened.

Step 4: Compare with the experimental value.
The actual (experimental) depression in freezing point is 7 K, which is less than 11.2 K.

Step 5: Find the van't Hoff factor (i).
$i = {7 \over 11.2} = 0.625$
(This value is less than 1. It shows association, not dissociation.)

Step 6: Find degree of association (α).
The formula is: $\alpha = {{i - 1} \over {{1 \over n} - 1}}$
where n = number of molecules coming together. For dimerization, n = 2.

Substitute values: $\alpha = {{0.625 - 1} \over {{1 \over 2} - 1}} = {{-0.375} \over {-0.5}} = 0.75$

Step 7: Calculate the percent dimerization.
The percentage is $0.75 \times 100 = 75\%$.

Answer: 75% of the phenol molecules dimerize in the solution.

(1) If molar mass of C₆H₅COOH is Mg mol^$-$1, then no. of moles of dissolved C₆H₅COOH = ${{1.22} \over M}$ mol

$\therefore$ Molality (m) of the solution = ${{1.22} \over M} \times {{1000} \over {100}} = {{12.2} \over M}$ mol kg^$-$1

We know, $\Delta$T_b = k_b $\times$ m

Given : $\Delta$T_b = 0.17 K and k_b = 1.7 K kg mol^$-$1

$\therefore$ $0.17 = 1.7 \times {{12.2} \over M}$ or, M = 122

$\therefore$ Molecular mass of C₆H₅COOH is = 122

(2) If molar mass of C₆H₅COOH is M', then no. of moles of dissolved C₆H₅COOH = ${{1.22} \over {M'}}$ mol

Molality (m) of the solution = ${{1.22} \over {M'}} \times {{1000} \over {100}} = {{12.2} \over {M'}}$ mol kg^$-$1

We know, $\Delta$T_b = 0.13 K and k_b = 2.6 K kg mol^$-$1

$\therefore$ $0.13 = 2.6 \times {{12.2} \over {M'}}$ or, M' = 244

$\therefore$ Molecular weight of C₆H₅COOH = 122

Now, van't Hoff factor (i) = ${{Calculated\,molecular\,weight} \over {Experimental\,molecular\,weight}} = {{122} \over {244}} = {1 \over 2}$

$i = {1 \over 2}$ indicates that C₆H₅COOH undergoes complete dimerisation in benzene.

Mass of water = 500 $\times$ 0.997 = 498.5 g and that of acetic acid = 3 $\times$ 10^$-$3 kg = 3 g.

No. of moles of acetic acid in solution = ${3 \over {60}} = 0.05$ mol [$\because$ Molar mass of acetic acid = 60 g mol^$-$1]

Molality (m) of the solution = ${{0.05} \over {498.5}} \times 1000 = 0.1$ mol kg^$-$1

Given : $\alpha$ = 0.23. Now, 1 molecule of CH₃COOH on dissociation in aqueous solution produces 2 ions,

CH₃COOH(aq) $\to$ CH₃COO^$-$(aq) + H⁺(aq)

So, n = 2

$\therefore$ $0.23 = {{i - 1} \over {2 - 1}}$ or, i = 1.23

For a solution of an electrolyte, the freezing point depression of the solution is given by, $\Delta$T_f = i $\times$ k_f $\times$ m

$\therefore$ $\Delta$T_f = 1.23 $\times$ 1.86 $\times$ 0.1 = 0.228 K

Hence, freezing point depression of solution = 0.228 K

As given in the question,

Determination of empirical formula of A :

Element with atomic number	Percentage	Relative Number	Simplest ratio
C (12)	42.86	${{42.86} \over {12}} = 3.57$	${{3.57} \over {1.19}} = 3.00$
H (1)	2.40	${{2.40} \over 1} = 2.40$	${{2.40} \over {1.19}} = 2.00$
N (14)	16.67	${{16.67} \over {14}} = 1.19$	${{1.19} \over {1.19}} = 1.00$
O (16)	38.07	${{38.07} \over {16}} = 2.37$	${{2.37} \over {1.19}} = 2.00$

$\therefore$ Empirical formula of minor product (A) = C₃H₂NO₂

Empirical formula mass = (12 $\times$ 3) + (1 $\times$ 2) + 14 $\times$ (16 $\times$ 2) = 84

Now, $\Delta$T_B = k_b $\times$ molality

or, $1.84 = 2.53 \times {{5.5} \over M} \times {1 \over {45}} \times 1000$

or, molar mass (M) = ${{2.53 \times 5.5 \times 1000} \over {1.84 \times 45}}$

$\therefore$ M = 168

$\therefore$ ${{molar\,mass} \over {formula\,mass}} = {{168} \over {84}} = 2$

$\therefore$ Molecular formula of (A) = (C₃H₂NO₂)₂ = C₆H₄N₂O₄

Structural formula :

We know, $\Delta$T_f = k_f $\times$ molality of solution (m)

Given, k_f for water = 1.86 K kg mol^$-$1

$\therefore$ 0.30 = 1.86 $\times$ molality of solution (m)

$\therefore$ m = ${{0.30} \over {1.86}}$ or, m = 0.161 mol kg^$-$1

According to Raoult's law, ${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$

Amount of water in solution n₁ $\times$ 18 g

Molality of the solution ${{{n_2}} \over {{n_1} \times 18}} \times 1000 = {{55.55 \times {n_2}} \over {{n_1}}}$ mol kg^$-$1

But, molality calculated from f. p. depression = 0.161 mol kg^$-$1

$\therefore$ ${{55.55 \times {n_2}} \over {{n_1}}} = 0.0161$ or, ${{{n_2}} \over {{n_1}}} = 1.09 \times {10^{ - 3}}$

Given : p⁰ = 23.51 mm Hg

$\therefore$ ${{23.51 - p} \over {23.51}} = {{{n_2}} \over {{n_1}}} = 1.09 \times {10^{ - 3}}$

$\therefore$ p = 23.48 mm Hg

Volume of 1 mol of benzene $ = {{78} \over {0.877}} = 88.94$ mL and that for 1 mol of toluene $ = {{92} \over {0.867}} = 106.11$ mL

At 20$^\circ$C, volume of 1 mol of benzene vapour $ = 88.94 \times 2750 = 244585$ mL = 244.58 L & that for 1 mol of toluene vapour $ = 106.11 \times 7720 = 819.16$ L

The vapour pressure for pure benzene, $p_B^0 = {{nRT} \over V} = {{1 \times 0.0821 \times 293} \over {244.58}}$ atm = 0.098 atm and that for toluene, $p_T^0 = {{nRT} \over V} = {{1 \times 0.0821 \times 293} \over {819.16}}$ atm = 0.029 atm

Again, total vapour pressure = VP of benzene $\times$ mole fraction of benzene + VP of toluene $\times$ mole fraction of toluene

or, $P = p_B^0 \times {x_B} + p_T^0 \times {x_T}$ ...... [1]

Since, ${x_B} + {x_T} = 1$ $\therefore$ ${x_T} = 1 - {x_B}$ [x_B, x_T are mole fractions of benzene & toluene, respectively]

Given, total vapour pressure = 46 torr $ = {{46} \over {760}}$ atm = 0.060 atm

From equation [1], we get, $0.060 = 0.098 \times {x_B} + 0.029(1 - {x_B})$

or, $0.060 = 0.098{x_B} - 0.029{x_B} + 0.029$

or, ${x_B} = 0.45$ (in liquid phase)

$\therefore$ ${x_T} = 1 - 0.45 = 0.55$ (in liquid phase)

$\therefore$ Mole fraction of benzene in vapour phase $ = {{p_B^0 \times {x_B}} \over P} = {{0.098 \times 0.45} \over {0.060}} = 0.735$

Given, mass of water (W₁) = 100 g.

Let, the vapour pressure of water = p⁰

As given, vapour pressure of urea solution (p) $ = ({p^0} - 0.25{p^0}) = 0.75{p^0}$

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {{{W_2}/{M_2}} \over {{W_1}/{M_1} + {W_2}/{M_2}}}$

or, ${{{p^0} - 0.75{p^0}} \over {{p^0}}} = {{{W_2}/60} \over {100/18 + {W_2}/60}}$ [molar mass of urea = 60 g mol^$-$1 & that of water = 18 g mol^$-$1]

or, ${1 \over 4} = {{{W_2}/60} \over {{{1000 + 3{W_2}} \over {180}}}}$ or, ${W_2} = 111.1$

$\therefore$ 111.1 g of urea needs to be dissolved in the solution.

No. of moles of dissolved urea $ = {{111.1} \over {60}} = 1.85$ mol

Molality (m) of the solution $ = {{1.85} \over {100}} \times 1000 = 18.5$ mol kg^$-$1

We know, $\alpha = {{i - 1} \over {n - 1}}$

where $\alpha$ = degree of dissociation of an electrolyte in solution, n = no. of ions produced by 1 molecule of the electrolyte on dissociation, i = van't Hoff factor.

In aqueous solution, $Ca{(N{O_3})_2}$ dissociates as:

$Ca{(N{O_3})_2}(aq) \to C{a^{2 + }}(aq) + 2NO_3^ - (aq)$

Therefore, n = 3 and given : $\alpha$ = 0.7

$\therefore$ $0.7 = {{i - 1} \over {3 - 1}}$ $\therefore$ $i = 2.4$

For an electrolyte solution, the relative lowering vapour pressure is given by,

${{{p^0} - p} \over {{p^0}}} = i \times {x_2} \approx i \times {{{n_2}} \over {{n_1}}} = 2.4 \times {{{n_2}} \over {{n_1}}}$

In solution, no. of moles of $Ca{(N{O_3})_2}({n_2}) = {7 \over {1.64}} = 0.042$ mol [$\therefore$ Molar mass of $Ca{(N{O_3})_2}$ = 164 g mol^$-$1] and that of water $({n_1}) = {{100} \over {18}} = 5.55$ mol

Given : p⁰ = 760 mm Hg

$\therefore$ ${{760 - p} \over {760}} = {{2.4 \times 0.042} \over {5.55}}$ $\therefore$ p = 746.2 mm Hg

$\therefore$ Vapour pressure of the solution = 746.2 mm Hg

Given, vapour pressure of pure benzene, p⁰ = 640 mm Hg, vapour pressure of the solution, p = 600 mm Hg, mass of benzene, W₁ = 39.0 g, mass of the solid, W₂ = 2.175 g

We know, M₁ = molar mass of benzene = 78.

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {{{W_2} \times {M_1}} \over {{M_2} \times {W_1}}}$

$\therefore$ ${{640 - 600} \over {640}} = {{2.175 \times 78} \over {{M_2} \times 39}}$ or, ${M_2} = {{2.175 \times 78 \times 640} \over {39 \times 40}}$

or, ${M_2} = 69.6$, i.e., molecular mass of the solid = 69.6.

Given, at 373 K vapour pressure of the aqueous solution of glucose, p = 750 mm Hg.

We know, boiling point of water = 373 K. At this temperature, vapour pressure of water, p0
= 760 mm Hg

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$

$\therefore$ ${{760 - 750} \over {760}} = {{{n_2}} \over {{n_1}}}$ or, ${n_2} = {n_1} \times 0.013$

(1) Amount of water in solution n₁ mol = 18 $\times$ n₁g

$\therefore$ Molality of the solution $ = {{{n_2} \times 1000} \over {18 \times {n_1}}} = {{{n_1} \times 0.013 \times 1000} \over {18 \times {n_1}}}$ = 0.72 mol kg^$-$1

(2) Mole fraction of solute in solution $ = {{{n_2}} \over {{n_1} + {n_2}}} = {{{n_1} \times 0.013} \over {{n_1} + {n_1} \times 0.013}} = 0.0128$

Total vapour pressure (P) of an ideal solution made up of ethanol and methanol is given by, $P = {x_1}p_1^0 + {x_2}p_2^0$ where $p_1^0$ & $p_2^0$ are the vapour pressures of pure ethanol and pure methanol respectively and x₁ and x₂ are the mole fractions of ethanol and methanol respectively in solution.

Now, number of moles of ethanol $ = {{60} \over {46}} = 1.3$ (Molar mass of ethanol = 46 g mol^$-$1

and that of methanol $ = {{40} \over {32}} = 1.25$ (Molar mass of methanol = 32 g mol^$-$1

Total moles of ethanol & methanol = (1.3 + 1.25) = 2.55 mol

$\therefore$ ${x_1} = {{1.3} \over {2.55}} = 0.50$ and ${x_2} = {{1.25} \over {2.55}} = 0.49$

Given : $p_1^0$ = 44.5 mm Hg and $p_2^0$ = 88.7 mm Hg

$\therefore$ $p = {x_1}p_1^0 + {x_2}p_2^0 = (0.50 \times 44.5 + 0.49 \times 88.7)$ mm Hg = 65.71 mm Hg

$\therefore$ Total vapour pressure of the solution = 65.71 mm Hg

mole fraction of methanol in vapour $ = {{Partial\,v.p.\,of\,methanol\,in\,vapour\,of\,the\,solution} \over {Total\,v.p.\,of\,solution}}$

$ = {{{x_2}p_2^0} \over P} = {{0.49 \times 88.7} \over {65.71}} = 0.66$

Given : ${C_x}{H_{2y}}{O_y} + 2x{O_2} \to xC{O_2} + y{H_2}O + x{O_2}$

After cooling, only CO₂ and O₂ will be present in gaseous state because water exists in liquid form at 0$^\circ$C and 1 atm.

$\therefore$ Number of moles of gases after cooling = x + x = 2x

Volume of gases after cooling = 2.24 L (given)

$\therefore$ Moles of gases after cooling $ = {{2.24} \over {22.4}} = 0.1$

$\therefore$ 0.1 = 2x, or, x = 0.05

As given in the question, water collected during cooling = 0.9 g

$\therefore$ Number of moles of water $ = {{0.9} \over {18}} = 0.05$; hence, y = 0.05

$\therefore$ The empirical formula of the organic compound is CH₂O.

Given, vapour pressure of pure water (p⁰) = 17.5 mm Hg

Lowering of vapour pressure (p⁰ $-$ p) = 0.104 mm Hg

No. of moles of water in solution $({n_1}) = {{1000} \over {18}} = 55.55$ mol of the molar mass of solute is Mg mol^$-$1, then the number of moles of solute in solution $({n_2}) = {{50} \over M}$ mol

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$

$\therefore$ ${{0.104} \over {17.5}} = {{50} \over {M \times 55.55}}$ or, $M = {{50 \times 17.5} \over {55.55 \times 0.104}}$ or, M = 151.4

$\therefore$ Molar mass of the organic compound = 151.4 g mol^$-$1

Now, empirical formula mass of CH₂O = 30 g mol^$-$1

$\therefore$ $n = {{Molecular\,mass} \over {Empirical\,formula\,mass}} = {{151.4} \over {30}}$ or, $n = 5.04 \simeq 5$

$\therefore$ Molecular formula $ = {[C({H_2}O)]_5} = {C_5}{H_{10}}{O_5}$

As given, vapour pressure of pure benzene, p⁰ = 639.7 mm Hg and that of solution, p = 631.9 mm Hg.

If, n₁ and n₂ be the number of moles of benzene and solute respectively, then according to Raoults' law,

${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$ or ${{639.7 - 631.9} \over {639.7}} = {{{n_2}} \over {{n_1}}}$

$\therefore$ ${n_2} = 0.0122 \times {n_1}$

Amount of benzene in solution = n₁ mol = 78 $\times$ n₁g [$\because$ Molar mass of benzene = 78 g mol^$-$1]

$\therefore$ Molality of the solution $ = {{{n_2} \times 1000} \over {78 \times {n_1}}} = {{0.0122 \times {n_1} \times 1000} \over {78 \times {n_1}}}$ = 0.156 mol kg^$-$1