# What is the molality when 48.0 mL of 6.00 M H2SO4 are diluted into 0.250 L?

Dec 22, 2014

The answer is $1.15 m$.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of ${H}_{2} S {O}_{4}$ and the mass of the solvent, which I presume is water.

We can find the number of ${H}_{2} S {O}_{4}$ moles by using its molarity

$C = \frac{n}{V} \to {n}_{{H}_{2} S {O}_{4}} = C \cdot {V}_{{H}_{2} S {O}_{4}} = 6.00 \frac{m o l e s}{L} \cdot 48.0 \cdot {10}^{- 3} L = 0.288$

Since water has a density of $1.00 \frac{k g}{L}$, the mass of solvent is

$m = \rho \cdot {V}_{w a t e r} = 1.00 \frac{k g}{L} \cdot 0.250 L = 0.250$ $k g$

Therefore, molality is

$m = \frac{n}{m a s s . s o l v e n t} = \frac{0.288 m o l e s}{0.250 k g} = 1.15 m$