What are the molar and molal concentrations of a #9.27*g# mass of sodium chloride dissolved in #30*mL# of water?

Question

Naomi Aronson · Accepted Answer

Well, we can work out the #"molality"#.

#"Molality"="Moles of solute"/"Kilograms of solvent"# #=((9.27*g)/(58.44*g*mol^-1))/(30*gxx10^-3*g*kg^-1)=5.29*mol*kg#... On the other hand.... #"Molarity"="Moles of solute"/"Litres of solution"# We expect the molarity to be near as dammit to the molality, even for this concentrated solution.....given that in this instance, volume of solution would be relatively unchanged with respect to the pure solvent.

Wyatt Adkins · Answer

undefined To find the molar concentration, first convert the mass of sodium chloride to moles using its molar mass, then divide by the volume of solution in liters. To find the molal concentration, divide the moles of solute by the mass of the solvent in kilograms.

1. Calculate moles of NaCl:
   Mass of NaCl = 9.27 g
   Molar mass of NaCl = 58.44 g/mol
   Moles of NaCl = 9.27 g / 58.44 g/mol = 0.159 mol

2. Convert volume of water to liters:
   Volume of water = 30 mL = 30/1000 L = 0.03 L

3. Calculate molar concentration:
   Molarity (M) = Moles of solute / Volume of solution in liters
                 = 0.159 mol / 0.03 L = 5.3 M

4. Calculate molal concentration:
   Mass of water = 30 mL = 30 g (since the density of water is 1 g/mL)
   Mass of water in kg = 30 g / 1000 = 0.03 kg
   Molality (m) = Moles of solute / Mass of solvent in kg
                = 0.159 mol / 0.03 kg = 5.3 m

So, the molar concentration of the solution is 5.3 M, and the molal concentration is 5.3 m.