Test Index

NCERT Class XII Chemistry
Chapter - Solutions
Questions with Solutions

© examsnet.com
Question : 36 of 53
Marks: +1, -0
100 g of liquid A (molar mass 140 g mol−1^{-1}) was dissolved in 1000 g of liquid B (molar mass 180 g mol−1^{-1}). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solutionif the total vapour pressure of the solution is 475 torr.
Solution:  
Number of moles of liquid A (solute) =100 g140 gmol−1=57 mole= \frac{100\ \mathrm{g}}{140\ \mathrm{gmol}^{-1}} = \frac{5}{7}\ \mathrm{mole}
Number of moles of liquid B (solvent) =1000 g180 g mol−1=509 mole= \frac{1000\ \mathrm{g}}{180\ \mathrm{g\ mol}^{-1}} = \frac{50}{9}\ \mathrm{mole}
∴ Mole fraction of A in the solution (xA)(x_A)
=5757+509= \frac{\frac{5}{7}}{\frac{5}{7} + \frac{50}{9}} =5739563= \frac{\frac{5}{7}}{\frac{395}{63}} =57×63395=45395=0.114= \frac{5}{7} \times \frac{63}{395} = \frac{45}{395} = 0.114
∴ Mole fraction of B in the solution (xB)=1−0.114=0.886(x_B) = 1 - 0.114 = 0.886
Also, given PB0=500P^{0}_B = 500torr
Applying Raoult’s law, PA=xAPA0=0.114×PA0P_A = x_A P^{0}_A = 0.114 \times P_A^0 ... (i)
PB=xBPB0=0.886×500=443P_B = x_B P_B^0 = 0.886 \times 500 = 443 torr
PTotal=PA+PBP_{\text{Total}} = P_A + P_B
475=0.114PA0+443475 = 0.114 P^{0}_A + 443 or PA0=475−4430.114=280.7P^{0}_A = \frac{475 - 443}{0.114} = 280.7 torr
Substituting this value in eqn. (i), we get PAP_A = 0.114 × 280.7 torr = 32 torr
© examsnet.com
Go to Question:
Prev Question
Next Question