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At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_b$ = 0.52, the boiling point of this solution will be
A
100°C
B
102°C
C
103°C
D
101°C
Explanation
Get moles of solute from RLVP, then apply ΔTb = Kb·m.
Detailed Solution
From Raoult's law: $\frac{p^o - p_S}{p_S} = \frac{n_2}{n_1} \Rightarrow \frac{760 - 732}{760} = \frac{n_2}{100/18}$
$n_2 = \frac{28\times100}{760\times18} = 0.2046$ moles
$\Delta T_b = K_b\times m = K_b\times\frac{n_2\times1000}{W(s)}$
$T_b - 100^\circ C = \frac{0.52\times0.2046\times1000}{100} = 1.06$
$T_b = 101.06^\circ C \approx 101^\circ C$
$n_2 = \frac{28\times100}{760\times18} = 0.2046$ moles
$\Delta T_b = K_b\times m = K_b\times\frac{n_2\times1000}{W(s)}$
$T_b - 100^\circ C = \frac{0.52\times0.2046\times1000}{100} = 1.06$
$T_b = 101.06^\circ C \approx 101^\circ C$