1. Tardigrade
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  3. Chemistry
  4. On mixing 3 g of non-volatile solute in 200 mL of water its boiling point (100° C ) becomes 100.52° C. If K b for water is 0.6 K / m then molecular weight of the solute is

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Q. On mixing $3\, g$ of non-volatile solute in $200\, mL$ of water its boiling point $\left(100^{\circ} C \right)$ becomes $100.52^{\circ} C$. If $K _{ b }$ for water is $0.6 \,K / m$ then molecular weight of the solute is

Solutions

Solution:

$\Delta T = K _{ b } \times \frac{ w }{ m } \times \frac{1000}{ W }$
$0.52=0.6 \times \frac{3}{ m } \times \frac{1000}{200}$
$m =\frac{1.8 \times 5}{0.52}=17.3 \,g \,mole ^{-1}$