$16.4 g$ of $Ca (NO X_{3}) X_{2}$ (molar mass $164 g mo l^{- 1}$ ) is dissolved in $90 g$ of water (molar mass $18 g mo l^{- 1}$ ). At the temperature of the experiment pure water has a vapour pressure of $760.0 mmHg$ and the solution has $722.4 mmHg$ . Using the exact mole-fraction form of Raoult's law with the effective number of solute particles, the degree of dissociation of $Ca (NO X_{3}) X_{2}$ is closest to

Answer: C

\frac{p ^{0} - p}{p ^{0}} = \frac{i n _{2}}{n _{1} + i n _{2}} = 0.049474

, so

i n_{2} = 0.049474 n_{1} / (1 - 0.049474)

. With

n_{1} = 5.00

and

n_{2} = 0.100

i = 2.602

. For

Ca (NO X_{3}) X_{2} Ca X^{2 +} + 2 NO X_{3} X^{-}

i = 1 + 2 α

, so

α = (2.602 - 1) /2 = 0.801 \approx 0.80

. Neglecting

i n_{2}

in the denominator gives

0.74

, which is not exact.