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# A certain non-volatile, non-electrolyte contains 40.0% carbon, 6.71% hydrogen and 53.3% oxygen. An aqueous solution containing 5% by mass of that solute which boils at 100.15$^{\large\circ}$C. Kb(water) = 0.512 K kg $mol^{-1}$. Calculate the molecular formula of that solute.

Hence the empirical formula is $CH_2O$. Empirical formula mass = 30g mol-1. Mass of solute = 5 g, mass of solvent = 100 – 5 = 95 g
$M_2=\large\frac{K_b W_2\times 1000}{W_1\times \Delta T_b}$
$\Rightarrow \large\frac{0.512\times W_2\times 1000}{95\times 0.15}$
$\Rightarrow 180gmol^{-1}$
Let the molecular formula is $(CH_2O)_x$
$x=\large\frac{\text{Molar mass}}{\text{Empirical mass}}=\frac{180}{30}$$=6$
Hence molecular formula = $C_6H_{12}O_6.$