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An aqueous solution of urea freezes at , for water = , for water = . The boiling point of urea…
An aqueous solution of urea freezes at $-0\cdot186^\circ\text{C}$, $K_f$ for water = $1\cdot86\text{ K kg mol}^{-1}$, $K_b$ for water = $0\cdot512\text{ K kg mol}^{-1}$. The boiling point of urea solution will be:
- (1)$373\cdot065\text{ K}$
- (2)$373\cdot186\text{ K}$
- (3)$373\cdot512\text{ K}$
- (4)$373\cdot0512\text{ K}$
Answer
Answer
AICorrect option: (4)
Answer: (4) $373\cdot0512\text{ K}$
$\Delta T_f = 0\cdot186\text{ K}$, so $m = \dfrac{0\cdot186}{1\cdot86} = 0\cdot1\text{ mol kg}^{-1}$
$\Delta T_b = 0\cdot512 \times 0\cdot1 = 0\cdot0512\text{ K}$, so boiling point $= 373 + 0\cdot0512 = 373\cdot0512\text{ K}$.
From ISC 2018 Chemistry Paper 1, question 1(b)(iii).