Question

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

Answer 1

The answer is `1.15m`.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of `H_2SO_4` and the mass of the solvent, which I presume is water.

We can find the number of `H_2SO_4` moles by using its molarity

`C = n/V -> n_(H_2SO_4) = C * V_(H_2SO_4) = 6.00(mol es)/L * 48.0 * 10^(-3) L = 0.288`

Since water has a density of `1.00 (kg)/L`, the mass of solvent is

`m = rho * V_(water) = 1.00 (kg)/L * 0.250 L = 0.250` `kg`

Therefore, molality is

`m = n/(mass.solvent) = (0.288 mol es)/(0.250 kg) = 1.15m`