Question

A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.4 ml . the density of water at 20.0∘c is 0.9982 g/ml.

Answer 1

The given question is incomplete. The complete question is as follows.

A 2.300×10−2 m solution of nacl in water is at 20.0∘c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.4 ml . the density of water at 20.0∘c is 0.9982 g/ml.

Calculate the molality of the salt solution.

Express your answer to four significant figures and include the appropriate units.

Explanation:

Molality is defined as the number of moles present in kg of a solvent.

Mathematically,     Molality = $\frac{\text{moles of solute}}{\text{mass of solvent}}$

Also,

      Mole of solute = Molarity of solute x Volume of solution

                               = (0.0230 M) x (1.000 L) = 0.0230 mol of solute

Therefore, mass of solvent will be as follows.

     $999.4 mL \times (\frac{0.9983 g}{1 mL})$

                  = 997.7 g

                  = 0.9977 kg        (as 1 kg = 1000 g)

Therefore, we will calculate the molality as follows.

          Molality = $\frac{0.0230 mol}{0.9977 kg}$

                   = 0.02306 mol/kg

thus, we can conclude that molality of the given solution is 0.02306 mol/kg.