Question

What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent

Answer 1

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

Explanation:

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

         $\text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}$

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, $\text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}$

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.