A 3.00 M aqueous solution of H2SO4 (M=98) has density 1.18 g mL^-1. Find its molality in mol kg^-1, to two decimal places.

JEE Advanced levelNumerical answerDifficulty 3 of 3mc161

A 3.00 M aqueous solution of $H X_{2} SO X_{4}$ ( $M = 98$ ) has density $1.18 g m L^{- 1}$ . Find its molality in $mol k g^{- 1}$ , to two decimal places.

Numerical answer type, answer in mol/kg, tolerance 0.01.

1 L of solution has mass 1180 g and contains 3 mol = 294 g of

H X_{2} SO X_{4}

. Water = 1180 - 294 = 886 g = 0.886 kg. Molality = 3/0.886 = 3.39 mol/kg.