1.80 g of a non-volatile solute in 90 g of benzene raises the boiling point from 353.23 K to 354.11 K. With Kb = 2.53 K kg/mol for benzene, the molar mass of the solute is:
A116 g/mol
B29 g/mol
C58 g/mol
D44 g/mol
Answer & Solution
Correct answer: C. 58 g/mol
1. $\Delta T_b = 354.11 - 353.23 = 0.88$ K.
2. $M_2 = (1000 \times w_2 \times K_b)/(\Delta T_b \times w_1)$.
3. = $(1000 \times 1.8 \times 2.53)/(0.88 \times 90)$.
4. = $4554 / 79.2 = 58$ g/mol. (Option B halves it, the error if $\Delta T_b$ were doubled.)
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.16_
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