(Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol-1 and 5.12 K kg mol-1, respectively).
Explanation:
Given, vapour pressure of solvent (p$^\circ $) = 650 mmHg
Vapour pressure of solution (ps) = 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}} $
$\Rightarrow \Delta {T_f} = 1.02K$
(given that the vapor pressure of pure liquid $A$ is $20$ $Torr$ at temperature $T$)
Explanation:
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
${{{\chi _A}} \over {{\chi _B}}} = {{0.95} \over {0.05}} = 19$
On addition of equal number of moles of a non-volatile solute $S$ in equal amount (in $kg$) of these solvents, the elevation of boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $Y$, the degree of dimerization in solvent $X$ is ___________.
Explanation:
For solvent X, $\Delta$TbX = KbX m $\Rightarrow$ 2 = 2KbX m ..... (1)
For solvent Y, $\Delta$TbY = KbY m $\Rightarrow$ 1 = 2KbY 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
${{\Delta {T_{bX}}} \over {\Delta {T_{bY}}}} = {{\left( {1 - {\alpha \over 2}} \right){K_{bX}}m} \over {\left( {{{1.3} \over 2}} \right){K_{bY}}m}} \Rightarrow {1 \over 3} = {{2(2 - \alpha )} \over {1.3}} \Rightarrow \alpha = 0.05$
(Kf for water=1.86oC kg mol−1) is approximately :
(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
Explanation:
$ 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 qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OH and CH3(CH2)11 OSO$_3^ - $ Na+ at room temperature. The correct assignment of the sketches is
I : KCl
II : CH3OH
III : CH3(CH2)11 OSO$_3^ - $ Na+
I. CH3(CH2)11 OSO$_3^ - $ Na+
II. CH3OH
III. KCl
I. KCl
II. CH3(CH2)11 OSO$_3^ - $ Na+
III. CH3OH
I. CH3OH
II. KCl
III. CH3(CH2)11 OSO$_3^ - $ Na+
Explanation:
The depression in freezing point is given by
$\Delta$Tf = Kf $\times$ 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.
Explanation:
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.
The Henry's law constant for the solubility of N$_2$ gas in water at 298 K is 1.0 $\times$ 10$^5$ atm. The mole fraction of N$_2$ in air is 0.8. The number of moles of N$_2$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is
The freezing point of the solution M is :
The vapour pressure of the solution M is :
Water is added to the solution M such that the fraction of water in the solution becomes 0.9 mole. The boiling point of this solution is:
When 20 g of naphthoic acid (C$_{11}$H$_{8}$O$_{2}$) is dissolved in 50 g of benzene in 50 g of benzene (K$_f$ = 1.72 K kg mol$^{-1}$), a freezing point depression of 2 K is observed. The van't Hoff factor (i) is :