Question
Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and ${PQ}_{2}$. When 1 g of PQ is dissolved in 50 g of solvent ' A '. $\Delta T_{b}$ was 1.176 K while when 1 g of ${PQ}_{2}$ is dissolved in 50 g of solvent ' A ', $\Delta T_{b}$ was 0.689 K . ( $K_{b}$ of ' A ' $= 5\text{ }Kkg{mol}^{- 1}$ ). The molar masses of elements P and Q (in $g{mol}^{- 1}$ ) respectively, are :
Options
Solution
$\left( \Delta T_{b} \right)_{PQ} = K_{b}m$
$${1.176 = 5 \times \frac{1}{M_{1}} \times \frac{1000}{50} }{M_{1} = 85.03 }{\left( \Delta T_{b} \right)_{{PQ}_{2}} = 5 \times \frac{1}{M_{2}} \times \frac{1000}{50} = 0.689 }{M_{2} = 145.13 }$$Let molar mass of $P\& Q$ are $M_{P}$ and $M_{Q}$ respectively
$${M_{P} + M_{Q} = 85.03 }{M_{P} + 2M_{Q} = 145.13 }{M_{p} = 24.93 \approx 25 }{M_{Q} = 60.1 \approx 60}$$
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