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NCERT Class XII Chemistry
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Question : 36
Total: 53
100 g of liquid A (molar mass 140 g mol‌–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol‌–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) =
100g
140g‌mol−1
=
5
7
 ‌mole
Number of moles of liquid B (solvent) =
1000g
180gmol−1
=
50
9
‌mole
∴ Mole fraction of A in the solution (xA)
=
5
7
5
7
+
50
9
 =
5
7
395
63
 =
5
7
×
63
395
=
45
395
=0.114
∴ Mole fraction of B in the solution (xB)=1–0.114=0.886
Also, given P0B=500torr
Applying Raoult’s law, PA=xAPA0=0.114×PA0 ... (i)
PB=xBPB0=0.886×500=443 torr
PTotal=PA+PB
475=0.114P0A+443 or P0A=
475−443
0.114
=280.7 torr
Substituting this value in eqn. (i), we get PA = 0.114 × 280.7 torr = 32 torr
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