Question

# A solution containing $$0.122\:kg$$ of benzoic acid in $$1\:kg$$ of benzene (b.pt.353 K) boils at $$354.5 K$$. Determine the apparent molar mass of benzoic acid (which dimerizes) in the solution and the degree of dimerization.Given: $$\:\Delta_{vap}H_{1m}\:(benzene)\,=\,394.57\:J\:g^{-1}$$.

A
0.214kgmol1,0.86
B
0.257kgmol1,0.932
C
0.428kgmol1,0.956
D
0.454kgmol1,0.986

Solution

## The correct option is A $$\;0.214\:kg\:mol^{-1},\,0.86$$In benzene, $$\displaystyle k_b = \frac {M_1RTT_b^2}{\Delta _{vap} H } = \frac {0.078 \times 8.314 \times (353)^2}{394.57 \times 78} = 2.626 K kg /mol$$The molality of the solution is $$\displaystyle m = \frac {\Delta T_b}{K_b} = \frac {354.5-353}{2.626} = 0.57 mol/kg$$$$\displaystyle M_2 = \frac {m_2}{mm_1} = \frac {0.122}{0.57 \times 1} = 0.214$$Let a be the degree of dimerization of benzoic aicd.The solution contains 1 mole of benzoic acid.$$\displaystyle \underset {(1-a) mol}{PhCOOH} \rightleftharpoons \underset {(a/2) mol}{(PhCOOH)_2 }$$Total amounts of solutes $$\displaystyle = (1-a + a/2) mol = (1-a/2) mol$$The number average molar mass of solutes in solution is $$\displaystyle \frac {m}{n_1+n_2} = \frac {0.122kg}{(1-a/2)mol}$$Equate this to 0.214 kg/mol$$\displaystyle \frac {0.122}{1-a/2} = 0.214$$$$\displaystyle a = 2(1-0.122/0.214) = 0.86$$Chemistry

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