Respuesta :
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 = [tex]\frac{\text{moles of solute}}{\text{mass of solvent}}[/tex]
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.
[tex]999.4 mL \times (\frac{0.9983 g}{1 mL})[/tex]
= 997.7 g
= 0.9977 kg (as 1 kg = 1000 g)
Therefore, we will calculate the molality as follows.
Molality = [tex]\frac{0.0230 mol}{0.9977 kg}[/tex]
= 0.02306 mol/kg
thus, we can conclude that molality of the given solution is 0.02306 mol/kg.
