Chemistry Class 12 NCERT Solutions: Chapter 2 Solutions Part 10

Q: 19. A solution containing Equation of non-volatile solute exactly in Equation of water has a vapour pressure of Equation at 298 K. Further, Equation of water is then added to the solution and the new vapour pressure becomes Equation at Equation . Calculate:

i. Molar mass of the solute

ii. Vapour pressure of water at 298 K.

Answer:

(i) Let, the molar mass of the solute be Equation

Now, the no. of moles of solvent (water), Equation

And, the no. of moles of solute, Equation

Equation

Applying the relation:

Equation

Equation

Equation

Equation

Equation

Equation

Equation

Equation …………………(i)

After the addition of 18 g of water:

Equation

Equation

Again, applying the relation

Equation

Equation

Equation

Equation

Equation

Equation

Equation

Equation …………………..(ii)

Dividing equation (i)by (ii), we have:

Equation

Equation

Equation

Equation

Equation

Equation

Therefore, the molar mass of the solute is Equation .

(ii) Putting the value of 'M' in equation (i), we have:

Equation

Equation

Equation

Hence, the vapour pressure of water at Equation .

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