1. Use Raoult’s law for an ideal solution

For liquids $X$ and $Y$:

$P_{\text{total}} = x_X P_X^0 + x_Y P_Y^0$

where $P_X^0$ and $P_Y^0$ are vapour pressures of pure $X$ and pure $Y$.

2. First solution: 1 mol of $X$ and 3 mol of $Y$

Total moles:

$n_{\text{total}} = 1+3=4$

So mole fractions are:

$x_X = \frac{1}{4}, \qquad x_Y = \frac{3}{4}$

Given total vapour pressure is $550$ mm Hg:

$\frac{1}{4}P_X^0 + \frac{3}{4}P_Y^0 = 550$

Multiplying by 4:

$P_X^0 + 3P_Y^0 = 2200 \qquad ...(1)$

3. After adding 1 more mol of $Y$

Now moles are:

$n_X=1, \qquad n_Y=4$

Total moles:

$n_{\text{total}}=5$

New mole fractions:

$x_X = \frac{1}{5}, \qquad x_Y = \frac{4}{5}$

Vapour pressure increases by $10$ mm Hg, so new total pressure is:

$550 + 10 = 560 \text{ mm Hg}$

Thus,

$\frac{1}{5}P_X^0 + \frac{4}{5}P_Y^0 = 560$

Multiplying by 5:

$P_X^0 + 4P_Y^0 = 2800 \qquad ...(2)$

4. Solve the two equations

From (2) $-$ (1):

$(P_X^0 + 4P_Y^0) - (P_X^0 + 3P_Y^0) = 2800 - 2200$

$P_Y^0 = 600 \text{ mm Hg}$

Substitute into (1):

$P_X^0 + 3(600) = 2200$

$P_X^0 + 1800 = 2200$

$P_X^0 = 400 \text{ mm Hg}$

5. Match with options

Thus,

$P_X^0 = 400 \text{ mm Hg}, \qquad P_Y^0 = 600 \text{ mm Hg}$

This corresponds to Option C.

6. Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.