Question

A solution is prepared by dissolving 400 g of NaOH in water and then diluting to one liter, The density of the resulting solution is 1.31 g/ml. Express the concentration of NaOH as Smiley: Angel percentage by weight (b) molarity Smiley: Coffee Cup normalty (d) molality Smiley: E Mail mole fraction

Answer 1

Answer on Question #39041-Chemistry-Organic Chemistry

Question

A solution is prepared by dissolving 400 g of NaOH in water and then diluting to one liter. The density of the resulting solution is 1.31 g/ml. Express the concentration of NaOH as

(a) percentage by weight

(b) molarity

(c) normality

(d) molality

Solution

Given:

m = 400 g – mass of the solute,

V = 1L = 1000 mL – volume of the solution,

ρ = 1.31 g/ml – density of the solution,

M = 40.0 g/mol – molar mass of the solute.

(a) Mass of the solution: $m_{s} = V \cdot \rho = 1000 \cdot 1.31 = 1310\, g = 1.31\, kg$

Percentage by weight:

$$w t.\% = \frac{m}{m_{s}}\ 100\% = \frac{400 \cdot 100}{1310} = 30.5\ \%_{by wt.}$$

(b) Molarity:

$$C_{M} = \frac{n}{V} = \frac{m}{M \cdot V} = \frac{400}{40.0 \cdot 1} = 10\, M$$

(c) Equivalence factor for NaOH – $f_{eq} = 1$, so the normality:

$$C_{N} = \frac{C_{M}}{f_{eq}} = \frac{10}{1} = 10\, N$$

(d) Mass of solvent:

$$m_{sol} = m_{s} - m = 1310 - 400 = 910\, g = 0.91\, kg$$

Molarity:

$$b = \frac{n}{m_{sol}} = \frac{m}{M \cdot m_{sol}} = \frac{400}{40.0 \cdot 0.91} = 11.0\, mol/kg$$

Answer: (a) 30.5 %by wt., (b) 10 M, (c) 10 N, (d) 11.0 mol/kg.