Molar mass of a non-volatile solute ( in g/mol) in its 10% by mass aqueous solution showing 2% decrease in its vapour pressure is :

100

The correct answer is option 1. 100.

We know,

The solution is 10% by mass in a non-volatile solute, meaning:

Out of 100 g of the solution, 10 g is the solute and 90 g is the solvent (water).

The vapor pressure of the solution is 2% lower than that of pure water. This means the relative lowering of vapor pressure is 2% or \(0.02\).

We need to find the molar mass of the solute.

Relative Lowering of Vapor Pressure

For solutions with a non-volatile solute, the presence of the solute lowers the vapor pressure of the solvent. Raoult's Law describes this effect, and the relative lowering of vapor pressure is given by:

\(\frac{\Delta P}{P_0} = \frac{n_2}{n_1}\)

where:

\(\Delta P = P_0 - P\) is the decrease in vapor pressure due to the solute,

\(P_0\) is the vapor pressure of the pure solvent (water in this case),

\(n_2\) is the number of moles of solute,

\(n_1\) is the number of moles of solvent.

However, when dealing with masses and molar masses, we can rewrite this in terms of the mass of solute (\(w_2\)), mass of solvent (\(w_1\)), molar mass of solvent (\(M_1\)), and molar mass of solute (\(M_2\))

\(\frac{\Delta P}{P_0} = \frac{w_2 \times M_1}{w_1 \times M_2}\)
where:

\(w_2\) is the mass of the solute,

\(M_1\) is the molar mass of the solvent,

\(w_1\) is the mass of the solvent,

\(M_2\) is the molar mass of the solute (which we are solving for).

Relative lowering of vapor pressure \(\frac{\Delta P}{P_0} = 0.02\) (since the vapor pressure decreased by 2%).

Mass of solute \(w_2 = 10\) g.

Mass of solvent \(w_1 = 90\) g.

Molar mass of solvent \(M_1 = 18\) g/mol (since water is the solvent, and water’s molar mass is 18 g/mol).

Using the formula for relative lowering of vapor pressure:

\(\frac{\Delta P}{P_0} = \frac{w_2 \times M_1}{w_1 \times M_2}\)

Plugging in the values we know:

\(0.02 = \frac{10 \times 18}{90 \times M_2}\)

or, \(M_2 = \frac{10 \times 18}{0.02 \times 90}\)

or, \(M_2 = 100 \, \text{g/mol}\)

The molar mass of the non-volatile solute is 100 g/mol.