Q.A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

Concept: Relative lowering of vapour pressure (Raoult's law for a non-volatile solute).

The same solute is present in two solutions of different dilution, and each solution's vapour pressure is known — so NCERT's relative-lowering relation (eq. 1.28), $\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1} = \frac{w_2 M_1}{M_2 w_1}$, applied to both solutions gives two equations in the two unknowns ($M_2$ and $p_1^0$).

1. Set up both equations (solute: $w_2 = 30$ g; water: $M_1 = 18$ g/mol):

   Solution 1 (90 g water, $p = 2.8$ kPa): $\quad 1 - \frac{2.8}{p_1^0} = \frac{30 \times 18}{M_2 \times 90} = \frac{6}{M_2}$

   Solution 2 (90 + 18 = 108 g water, $p = 2.9$ kPa): $\quad 1 - \frac{2.9}{p_1^0} = \frac{30 \times 18}{M_2 \times 108} = \frac{5}{M_2}$

2. Subtract the second from the first — the 1's cancel: …

Concept: Relative Lowering of Vapour Pressure

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the resulting solution is lower than that of the pure solvent: solute particles occupy part of the liquid surface, reducing the number of solvent molecules that can escape into the vapour phase. Raoult's law quantifies this, and for a dilute solution NCERT expresses it as the relative lowering of vapour pressure (eq. 1.28):

Here the same 30 g of solute appears in two solutions of different dilution, each with a measured vapour pressure. That gives two equations sharing the same two unknowns — the solute's molar mass $M_2$ and pure water's vapour pressure $p_1^0$ — which we solve simultaneously.

Step-by-Step Solution

1. Write the relation for the first solution.

Mass of water $w_1 = 90$ g, solution vapour pressure $p_1 = 2.8$ kPa:

which rearranges to

2. Write the relation for the second solution.

After adding 18 g of water, the total water is $90 + 18 = 108$ g, and the vapour pressure is 2.9 kPa:

3. Subtract Equation 2 from Equation 1.

The 1's cancel, leaving:

4. Substitute back to find $M_2$.

Putting $p_1^0 = 0.1\,M_2$ into Equation 1:

5. Find $p_1^0$. …

Method: Two-State Relative Lowering of Vapour Pressure

The same solute is dissolved twice at different dilutions, and each state's vapour pressure is measured. Applying the relative-lowering relation (NCERT eq. 1.28) to each state gives a system of two equations in the two unknowns — the solute's molar mass $M_2$ and the pure solvent's vapour pressure $p_1^0$.

Step 1: Write the relation

For water, $M_1 = 18$ g/mol; the solute mass is $w_2 = 30$ g throughout.

Step 2: Apply it to each solution

Solution 1 — $w_1 = 90$ g water, $p_1 = 2.8$ kPa:

Solution 2 — $w_1 = 90 + 18 = 108$ g water, $p_1 = 2.9$ kPa:

Step 3: Eliminate one unknown by subtraction

Subtracting the second equation from the first (the 1's cancel):

Step 4: Substitute back …

Here are the common mistakes students make on this exact type of problem (finding the molar mass of a solute AND the solvent's own vapour pressure from two solutions of different concentration), and how to avoid each.

1. Forgetting the Total Water Mass After Dilution

The Mistake: Using 18 g (the amount added) instead of 108 g (90 g original + 18 g added) as the new solvent mass in the second equation.

How to avoid: Read carefully — "18 g of water is then added to the solution" means the new total water mass is $90+18=108$ g, not 18 g.

2. Trying to Solve With Only One Equation

The Mistake: Using only the first solution's data (90 g water, 2.8 kPa) and assuming a "standard" memorised value for the vapour pressure of pure water instead of treating it as unknown.

Why it's wrong: The problem deliberately gives two solutions so that you can solve for two unknowns ($M_2$ and $p_1^0$) simultaneously — plugging in a memorised value for $p_1^0$ defeats the purpose and gives an inconsistent answer.

How to avoid: Count the unknowns the problem has, and set up exactly that many independent equations.

3. Setting Up the Relative-Lowering Relation Inconsistently

The Mistake: Writing the lowering over the solution pressure, $\frac{p_1^0 - p_1}{p_1}$, in one equation and over the pure-solvent pressure, $\frac{p_1^0 - p_1}{p_1^0}$, in the other — or mixing $\frac{n_2}{n_1}$ with $\frac{n_2}{n_1+n_2}$ between the two equations.

Why it matters here: NCERT's eq. 1.28 uses $\frac{p_1^0 - p_1}{p_1^0} = \frac{n_2}{n_1}$, and this problem's expected answers (34 g/mol, 3.4 kPa) come from applying exactly that form to both states. Because this solution is fairly concentrated, sloppy mixing of forms changes the numbers noticeably — consistency is everything.

How to avoid: Write the formula once at the top, then substitute into it twice. Use the book's form (eq. 1.28) throughout.

4. Arithmetic Slips in the Elimination …