Question

30grams of non volatile solute is dissolved in 360grams of water at 100 degree centigrade if the VP of solution is 570millimetres of mercury then calculate molar mass of solute​

Answer 1

Molar mass of the solute is 45 g.

Given,

Mass of the non-volatile solute= 30 g

Mass of water=360 g

Vapour pressure of solution=570 mm of Hg

Temperature=100 °C.

To find,

the molar mass of the solute.

Solution:

• The property referred to here is the relative lowering of vapour pressure.
• It is a colligative property and it only depends on the amount of solute in the solution.
• In presence of non-volatile solute, the vapour pressure of solution is less than the vapour pressure of the pure solvent.
• Relative lowering of vapour pressure is equal to the mole fraction of the solute.

$\frac{Po-Ps}{Po} =Xa$

where,

Po-vapour pressure of pure solvent

Ps-vapour pressure of solution

Xa-mole fraction of non-volatile solute

$Xa=\frac{na}{na+nb}$

where,

na-moles of solute

nb-moles of solvent

As the solution here is at its boiling point i.e. 100°C so ,

$Po=1 atm\\Po=760 mm of Hg$

Now,

$\frac{Po-Ps}{Po} =Xa\\\frac{760-570}{760}=Xa\\Xa=\frac{190}{760} \\Xa=\frac{1}{4}$

The mole fraction of non-volatile solute comes as $\frac{1}{4}$.

$Xa=\frac{na}{na+nb}$

Moles=Given mass/Molar mass

Molar mass of water= 18 g.

Let the molar mass of the non-volatile solute be $x$ g.

$\frac{1}{4} =\frac{\frac{30}{x} }{\frac{30}{x}+\frac{360}{18} } \\\frac{1}{4} =\frac{\frac{30}{x} }{\frac{30}{x}+20}\\\frac{1}{4}=\frac{\frac{30}{x} }{\frac{30+20x}{x}}\\\\\frac{1}{4}=\frac{30}{30+2x} \\30+2x=4X30\\30+2x=120\\2x=120-30\\2x=90\\x=45 g.$

The molar mass of the solute comes as 45 g.

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