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# Calculate the molality of 1 litre solution of 93% (w/v) $H_2SO_4$. The density of solution is 1.84 g/ml.

Weight volume percentage = w-v% = 93 g/ml
Gram molecular mass of H2SO4 = GMM = 98 g
Density of solution = d = 1.84 g/ml
Molality = $\large\frac{w}{GMM}\times \frac{1000}{\text{Weight of solvent in grams}}$
$m=\large\frac{93}{98}\times \frac{1000}{91}$$=10.42moles/kg$
Hence (A) is the correct answer.