﻿ The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a one molal solution of a non-volatile, non-ionic solute in water. from Chemistry Solutions Class 12 Nagaland Board

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The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a one molal solution of a non-volatile, non-ionic solute in water.

Solution
Step-1 what is given in problem
Vapour pressure of pure water = 12.3 kPa
We have 1 molal solution that means we have 1 moles of solute in 1kg of solvent
Step – 2 Find vapour pressure
The formula of vapour pressure when non-volatile liquid is added
Pressure = vapour pressure of pure liquid × molar fraction of liquid

P=P(pure)X  ...........1

Here we know the value of P(pure)so need to find the value of molar fraction X

Here we know that number of moles of solute =1 moles
And need to find the moles of solvent (water)

And we have mass = 1 kg and 1 kg is equal to 1000g
And molar mass of water H2O = 2×1 + 16 = 18 g/mol
Plug the value in equation (3) we get
Number of moles of water =1000 g / (18g/mol  )  =55.56 moles
Plug the value in equation (2) we get
The formula of molar fraction =55.5/(55.5+1)  = 0.9823
Now plug in equation (1 )we get
Partial fraction = 12.3×0.9823 = 12.08

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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 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 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 (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

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