Question

A 560.0 mL of 0.61 M sodium bromide, NaBr solution has a density of 1.046 g mL-1

. Calculate the

molality of the solution.

Answer 1

$V=560.0 \;mL \\ ρ=1.046 \;g/mL \\ m(solution) = ρV = 1.046 \times 560.0 = 585.76 \;g$

Number of moles of NaBr present in the solution:

Proportion:

0.61 mol – 1000 mL

x mol – 560 mL

$x = \frac{0.61 \times 560}{1000}= 0.3416 \;mol$

Number of moles of NaBr present in the solution is 0.3416

$M(NaBr) = 102.894 \;g/mol \\ m(NaBr) = 0.3416 \times 102.894 = 35.148 \;g$

Mass of water in the solution $= (585.76 -35.148) = 550.612 \;g$

Molality of the NaBr solution $= \frac{0.3416}{550.612} \times 1000 = 0.6204 \;m$