Question

4L of 0.02 M aqueous solution of NaCl was diluted by adding one liter of water. The molality of the resultant solution is:

a. 0.01 m
b. 0.02 m
c. 0.04 m
d. 0.08 m

Answer 1

The molality of the resultant solution is 0.004 m.

Step-by-step explanation:

The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, we have 4L of 0.02 M NaCl solution, which means we have 0.02 moles of NaCl. To calculate the molality, we need to convert the mass of the solvent to kilograms. Since the density of water is 1 g/mL, 1 liter of water has a mass of 1000 grams or 1 kilogram. Adding 1 liter of water gives us a total mass of 5 kilograms. Therefore, the molality of the resultant solution is 0.02 mol / 5 kg = 0.004 mol/kg, which is equivalent to 0.004 m.