Embibe Experts Solutions for Chapter: Solutions, Exercise 1: NEET - 24th July 2016

The following solutions were prepared by dissolving $10 \mathrm{g}$ of glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ in $250 \mathrm{mL}$ of water $\left({\mathrm{P}}_1\right),10 \mathrm{g}$ of urea $\left({\mathrm{CH}}_4{\mathrm{N}}_2\mathrm{O}\right)$ in $250 \mathrm{mL}$ of water $\left({\mathrm{P}}_2\right)$ and $10 \mathrm{g}$ of sucrose $\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ in $250 \mathrm{mL}$ of water $\left({\mathrm{P}}_3\right).$ The right option for the decreasing order of osmotic pressure of these solutions is:

The correct option for the value of vapour pressure of a solution at $45º\mathrm{C}$ with benzene to octane in molar ratio $3 : 2$ is:

[At $45°\mathrm{C}$ vapour pressure of benzene is $280 \mathrm{mm} \mathrm{Hg}$ and that of octane is $420 \mathrm{mm} \mathrm{Hg}.$ Assume Ideal gas]

If $8 \mathrm{g}$ of a non-electrolyte solute is dissolved in $114 \mathrm{g}$ of $\mathrm{n}-$octane to reduce its vapour pressure to $80\%,$ the molar mass (in $\mathrm{g} {\mathrm{mol}}^{-1}$) of the solute is [Given that molar mass of n-octane is $114 \mathrm{g} {\mathrm{mol}}^{-1}$]

Isotonic solutions have same

The freezing point depression constant $\left({\mathrm{K}}_{\mathrm{f}}\right)$ of benzene is $5.12\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{–1}$. The freezing point depression for the solution of molality $0.078\mathrm{m}$ containing a non-electrolyte solute in benzene is (rounded off upto two decimal places):

The mixture which shows positive deviation from Raoult’s law is

The mixture that forms maximum boiling azeotrope is

Of the following $0.10\mathrm{m}$ aqueous solutions, which one will exhibit the largest freezing point depression?