12 g of a nonvolatile solute dissolved in 108 g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of the solute is :

We are given that 12 grams of a nonvolatile solute is dissolved in 108 grams of water, and the relative lowering of vapour pressure is 0.1. We need to find the molecular mass of the solute.

The relative lowering of vapour pressure is given by the formula:

$$\frac{\Delta P}{P^0} = \frac{n_2}{n_1 + n_2}$$

where $$\Delta P = P^0 - P$$ is the lowering of vapour pressure, $$P^0$$ is the vapour pressure of pure solvent, $$P$$ is the vapour pressure of the solution, $$n_1$$ is the number of moles of solvent, and $$n_2$$ is the number of moles of solute.

For dilute solutions, since the solute is nonvolatile and the solution is dilute, we can approximate $$n_1 + n_2 \approx n_1$$ because $$n_2$$ is much smaller than $$n_1$$. So the formula simplifies to:

$$\frac{\Delta P}{P^0} \approx \frac{n_2}{n_1}$$

We are given that the relative lowering of vapour pressure is 0.1, so:

$$\frac{\Delta P}{P^0} = 0.1$$

Therefore:

$$0.1 = \frac{n_2}{n_1}$$

Now, we express $$n_1$$ and $$n_2$$ in terms of masses and molecular masses. Let $$M_2$$ be the molecular mass of the solute (which we need to find). The molecular mass of water (solvent) is known to be 18 g/mol.

So, the number of moles of solvent (water) is:

$$n_1 = \frac{\text{mass of water}}{\text{molecular mass of water}} = \frac{108}{18}$$

Calculating that:

$$n_1 = \frac{108}{18} = 6$$

The number of moles of solute is:

$$n_2 = \frac{\text{mass of solute}}{\text{molecular mass of solute}} = \frac{12}{M_2}$$

Substituting these values into the equation $$0.1 = \frac{n_2}{n_1}$$:

$$0.1 = \frac{\frac{12}{M_2}}{6}$$

Dividing by 6 is the same as multiplying by $$\frac{1}{6}$$:

$$0.1 = \frac{12}{M_2} \times \frac{1}{6}$$

Simplifying the right side:

$$0.1 = \frac{12}{6 \times M_2} = \frac{2}{M_2}$$

Now, solve for $$M_2$$:

$$0.1 = \frac{2}{M_2}$$

Multiply both sides by $$M_2$$:

$$0.1 \times M_2 = 2$$

Divide both sides by 0.1:

$$M_2 = \frac{2}{0.1}$$

Calculating the division:

$$M_2 = 20$$

Therefore, the molecular mass of the solute is 20 g/mol.

Looking at the options:

A. 80

B. 60

C. 20

D. 40

Hence, the correct answer is Option C.