Question

Lowering in vapor pressure is highest for:

• A. $0.2\mathrm{M}\mathrm{Urea}$
• B. $0.1\mathrm{M}{\mathrm{BaCl}}_2$
• C. $0.1\mathrm{M}{\mathrm{MgSO}}_4$
• D. $0.1\mathrm{M}\mathrm{Glucose}$

Answer 1

The correct option is B

$0.1\mathrm{M}{\mathrm{BaCl}}_2$

The correct option is (B):-

Explanation for the correct answer:

Step 1: Finding the relation between the Lowering of Vapor Pressure and Van't Hoff factor (i).

$\frac{\mathrm{P}°-{\mathrm{P}}_{\mathrm{S}}}{\mathrm{P}°}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{W}}_{\mathrm{A}}}\times {\mathrm{M}}_{\mathrm{A}}\times \mathrm{i}$

$\frac{\mathrm{P}°-{\mathrm{P}}_{\mathrm{S}}}{\mathrm{P}°}=\mathrm{m}\times {\mathrm{m}}_{\mathrm{A}}\times \mathrm{i}$

$∆\mathrm{P}=\mathrm{P}°-{\mathrm{P}}_{\mathrm{S}}=\mathrm{P}°\times \mathrm{m}\times {\mathrm{m}}_{\mathrm{A}}\times \mathrm{i}$

( where, $∆\mathrm{P}\mathrm{stands}\mathrm{for}\mathrm{relative}\mathrm{lowering}\mathrm{in}\mathrm{vapor}\mathrm{pressure}$ )

Therefore, $∆\mathrm{P}\mathrm{is}\mathrm{directly}\mathrm{proportional}\mathrm{to}\mathrm{m}\times \mathrm{i}$

Step 2: Calculating value of $\mathrm{m}\times \mathrm{i}$ for each.

For Urea, i = 1 (non-electrolyte) So, $\mathrm{m}\times \mathrm{i}=0.2\times 1=0.2$

For ${\mathrm{BaCl}}_2$, i = 3 (because of 2 ${\mathrm{Cl}}^{-}$ ions and 1 ${\mathrm{Ba}}^{2+}\mathrm{ion}$) So, $\mathrm{m}\times \mathrm{i}=0.1\times 3=0.3$

For ${\mathrm{MgSO}}_4$, i = 2 (because of 1 ${\mathrm{Mg}}^{2+}$ and 1 ${\mathrm{SO}}_4^{2-}$ ions) So, $\mathrm{m}\times \mathrm{i}=0.1\times 2=0.2$

For Glucose, i = 1(non-electrolyte) So, $\mathrm{m}\times \mathrm{i}=0.1\times 1=0.1$

Hence, $\mathrm{m}\times \mathrm{i}$ for ${\mathrm{BaCl}}_2$ is highest, so as $∆\mathrm{P}$ (lowering in vapor pressure)

Therefore, the correct option is (B):- $0.1\mathrm{M}{\mathrm{BaCl}}_2$