Answer to Question #166064 in Inorganic Chemistry for Wilhelm

Question

Given 0.2 M solution of NaCl in water with a specific gravity of 1.35.

• How many grams of NaCl are present in 800 mL of solution?
• What is the molality of the solution?

Answer 1

$Specific \; gravity = \frac{ρ_{NaCl}}{ρ_{water}} \\ 1.35 = \frac{ρ_{NaCl}}{1} \\ ρ_{NaCl} = 1.35 \;g/mL \\ ρ = \frac{m}{V} \\ m = ρ \times V \\ m_{solute} = 1.35 \times 800 = 1080 \;g = 1.08 \;kg$

Proportion:

0.2 mol – 1000 mL

x mol – 800 mL

$x =\frac{0.2 \times 800}{1000}=0.16 \;mol \\ m = n \times M$

M(NaCl) = 58.44 g/mol

$m(NaCl) = 0.16 \times 58.44 = 9.35 \;g$

9.35 grams of NaCl are present in 800 mL of solution.

The formula for molality is

m = moles of solute / kilograms of solvent

$= \frac{0.16 \;mol}{1.08 \;kg} \\ = 0.148 \;mol/kg$

0.148 mol/kg is the molality of the solution.