Question

The molecular weight of an organic compound is $58g{mol}^{-1}$. What will be the boiling point of a solution containing $48$ gram of the solute in $1200$ gram of water ?
[$K_b$ for water $=0.513kg{mole}^{-1}$; Boiling point of water $=100$.]

Answer 1

Molecular weight $=58gm/mol$,

Boiling point of solution $=$ ?

Solution : $48gm$ solute in $1200gm$ $H_{2}O$

$K_b$ for water $={0.513}^{o}C$

Boiling point of water $=100\left(T_b\right)$

$\Delta T_b=\frac{1000\times K_b\times w}{M\times W}$

$\Delta T_b=\frac{1000\times 0.513\times 58}{48\times 1200}$

$\Delta T_b=0.516$

$\Delta T_b=T-T_b$

$T=100.516$