﻿ 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 and found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr. from Chemistry Solutions Class 12 Nagaland Board

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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 and found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Solution
Given that
Mass of liquid A , WA = 100g
Molar mass, MA = 140 g mol – 1
Mass of liquid B, WB = 1000 g
Molar mass, MB = 180 g mol – 1
Use the formula

Number of moles of liquid A, MA = 100/140 = 0.714 mol
Number of moles of liquid B, MB =1000/ 180 = 5.556 mol
Use formula

Molar fraction of A,XA = 0.714 /(0.714 + 5.556) = 0.114
Similarly
Molar fraction of B, XB   = 1-  XA
= 1 − 0.114 = 0.886
Vapour pressure of pure liquid B, PoB = 500 torr
Use formula of Henrys law
PB  = Po× XB
Plug the values we get
PB= 500 × 0.886 = 443 torr
Given that total vapour pressure of the solution, ptotal = 475 torr
Use the formula
Ptotal   = pA + pB
pA      = ptotal − pB
Plug the values we get
PA      = 475 − 443
PA      = 32 torr
Use formula of Henrys law again we get
PA  = Po× XA
Plug the values we get
32 = PoA × 0.114
PoA = 32/0.114  = 280.7 torr
So that the vapour pressure of pure liquid A = 280.7 torr.

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Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,

897 Views

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Mass % of benzene

Mass% of carbon tetrachloride = 100 - 15.28
= 84.72%
1703 Views

Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.

solution;

Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

(a) Mol. mass of

Moles of $\mathrm{Co}\left(\mathrm{NO}{\right)}_{3}.6{\mathrm{H}}_{2}\mathrm{O}$

Volume of solution = 4.3 L
Molarity,

(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=$\frac{0.5×30}{1000}$mol =0.015mol
therefore molarity =0.015/0.5L

thus molarity is 0.03M

844 Views

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.

Solution:

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$
= 14 + 2 + 12 + 16 + 14 + 2
=

Molality (m) =

or Moles of solute
= 0.25 x 0.25 =  0.625

Mass of urea
= Moles of solute x Molar mass

= 0.625 x 60 = 37.5 g

1475 Views

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.

(a) 20% (mass/mass) means that 20 g of KI is present in 80 g of water.

Therefore, Moles of KI in solution

moles of KI = 20/166 =0.12mol
moles of water =80/18 =4.44mol
therefore, mole fraction of KI

=

1010 Views