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ChemHardAdv 2026 • Paper 2

Ratio of Molar Volume of Pure Component in Vapour to Liquid Phase

Comprehension Passage

Two volatile liquids A\textbf{A} and B\textbf{B} form an ideal solution. Consider a 5 molal5\text{ molal} solution of B\textbf{B} in A\textbf{A} inside a closed container having a total vapour pressure of 100 mm Hg100\text{ mm Hg} at 300 K300\text{ K}. The vapour pressure of pure A\textbf{A} at 300 K300\text{ K} is 105 mm Hg105\text{ mm Hg}. Assume that A\textbf{A} and B\textbf{B} behave as ideal gases in the vapour phase.

Given:\textbf{\small{Given:}}
The gas constant R=0.08 L atm K−1 mol−1R = 0.08\text{ L atm K}^{-1}\text{ mol}^{-1}
Molar mass of \textbf{A} is 50 g mol−150\text{ g mol}^{-1}
Molar mass of \textbf{B} is 57 g mol−157\text{ g mol}^{-1}
Density of liquid \textbf{B} at 300 K300\text{ K} is 0.5 g/mL0.5\text{ g/mL}
1 atm=760 mm Hg1\text{ atm} = 760\text{ mm Hg}

At 300 K300\text{ K}, the ratio of the molar volume of pure B\textbf{B} in vapour phase to its molar volume in liquid phase is _____.

Official Numerical Answer1991 to 2053

Topics & Concepts

SolutionsConcentration TermsIdeal Gas Law

Step-by-Step Solution

To find the ratio of the molar volume of pure B\mathbf{B} in the vapour phase to its molar volume in the liquid phase at 300 K300\text{ K}, we follow these steps:

Step 1: Calculate the mole fraction of liquid B (xBx_B) and liquid A (xAx_A)

A 5 molal5\text{ molal} solution contains 5 moles5\text{ moles} of solute B\mathbf{B} per 1000 g1000\text{ g} of solvent A\mathbf{A}.

  • Molar mass of A\mathbf{A}, MA=50 g mol−1M_A = 50\text{ g mol}^{-1}
  • Number of moles of A\mathbf{A}, nA=1000 g50 g mol−1=20 moln_A = \frac{1000\text{ g}}{50\text{ g mol}^{-1}} = 20\text{ mol}
  • Number of moles of B\mathbf{B}, nB=5 moln_B = 5\text{ mol}

The mole fractions in the liquid phase are: xA=nAnA+nB=2020+5=0.8x_A = \frac{n_A}{n_A + n_B} = \frac{20}{20 + 5} = 0.8 xB=nBnA+nB=520+5=0.2x_B = \frac{n_B}{n_A + n_B} = \frac{5}{20 + 5} = 0.2

Step 2: Determine the vapour pressure of pure B (PB∘P_B^\circ)

Using Raoult's law for ideal solutions: Ptotal=xAPA∘+xBPB∘P_{\text{total}} = x_A P_A^\circ + x_B P_B^\circ

Given Ptotal=100 mm HgP_{\text{total}} = 100\text{ mm Hg} and PA∘=105 mm HgP_A^\circ = 105\text{ mm Hg}: 100=(0.8×105)+(0.2×PB∘)100 = (0.8 \times 105) + (0.2 \times P_B^\circ) 100=84+0.2PB∘100 = 84 + 0.2 P_B^\circ 0.2PB∘=160.2 P_B^\circ = 16 PB∘=80 mm HgP_B^\circ = 80\text{ mm Hg}

In atmospheres: PB∘=80760 atmP_B^\circ = \frac{80}{760}\text{ atm}

Step 3: Calculate the molar volume of pure B in the vapour phase (Vm,vapV_{m, \text{vap}})

Assuming ideal gas behaviour for the vapour phase of pure B\mathbf{B} at its saturated vapour pressure PB∘P_B^\circ: PB∘Vm,vap=RTP_B^\circ V_{m, \text{vap}} = RT Vm,vap=RTPB∘V_{m, \text{vap}} = \frac{RT}{P_B^\circ}

Given R=0.08 L atm K−1 mol−1R = 0.08\text{ L atm K}^{-1}\text{ mol}^{-1} and T=300 KT = 300\text{ K}: Vm,vap=0.08×30080760=24×76080=228 L mol−1V_{m, \text{vap}} = \frac{0.08 \times 300}{\frac{80}{760}} = \frac{24 \times 760}{80} = 228\text{ L mol}^{-1}

Step 4: Calculate the molar volume of pure B in the liquid phase (Vm,liqV_{m, \text{liq}})

Given the density of liquid B\mathbf{B}, dB=0.5 g/mLd_B = 0.5\text{ g/mL} and its molar mass MB=57 g mol−1M_B = 57\text{ g mol}^{-1}: Vm,liq=MBdB=57 g mol−10.5 g mL−1=114 mL mol−1=0.114 L mol−1V_{m, \text{liq}} = \frac{M_B}{d_B} = \frac{57\text{ g mol}^{-1}}{0.5\text{ g mL}^{-1}} = 114\text{ mL mol}^{-1} = 0.114\text{ L mol}^{-1}

Step 5: Calculate the ratio

Ratio=Vm,vapVm,liq=228 L mol−10.114 L mol−1=2000\text{Ratio} = \frac{V_{m, \text{vap}}}{V_{m, \text{liq}}} = \frac{228\text{ L mol}^{-1}}{0.114\text{ L mol}^{-1}} = 2000

Thus, the ratio of the molar volume of pure B\mathbf{B} in the vapour phase to its molar volume in the liquid phase is 2000.

Ratio of Molar Volume of Pure Component in Vapour to Liquid Phase | Chemistry PYQ Solution - JEE Challenger