SolutionHard
Question
What is the boiling point of a solution of 1.00 g of naphthalene dissolved in 94.0 g of toluene? The normal boiling point of toluene is 110.75°C and Kx,b = 32.0 K for toluene.
Options
A.110.95°C
B.111.0°C
C.113.41°C
D.110.75°C
Solution
$\Delta T_{b} = K_{x,b}.X_{\text{solute}} = 32 \times \frac{1/128}{\frac{1}{128} + \frac{94}{94}} \approx 0.25^{o}C$
∴ B.P. of solution = 110.75 + 0.25 = 111° C
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