One mole of sugar is dissolved in two moles of water. The vapour pressure of the solution relative to that of pure water is:

2 / 3

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Solution

The correct option is D $2 / 3$
we have,

$P_{A} = P_{A}^{0} x_{A}$

$x_{A} = \frac{n_{A}}{n_{A} + n_{B}}$

$n_{A} = 2, n_{B} = 1$

$\frac{P_{A}}{P_{A}^{0}} = \frac{n_{A}}{n_{A} + n_{B}}$ = $\frac{2}{2 + 1}$ = $\frac{2}{3}$
Hence, option (A) is correct.

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One mole of sugar is dissolved in two moles of water. The vapour pressure of the solution relative to that of pure water is:

The correct option is D 2/3we have,PA=P0AxA xA=nAnA+nBnA=2,nB=1PAP0A=nAnA+nB = 22+1 = 23Hence, option (A) is correct.

The correct option is D $2 / 3$
we have,

$P_{A} = P_{A}^{0} x_{A}$

$x_{A} = \frac{n_{A}}{n_{A} + n_{B}}$

$n_{A} = 2, n_{B} = 1$

$\frac{P_{A}}{P_{A}^{0}} = \frac{n_{A}}{n_{A} + n_{B}}$ = $\frac{2}{2 + 1}$ = $\frac{2}{3}$
Hence, option (A) is correct.