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.

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3 years ago