Question

A 30.2 mL aliquot of a 30.0 wt% aqueous KOH solution is diluted to 1.20 L to produce a 0.173 M KOH solution. Calculate the density of the 30.0 wt% KOH solution.

Answer 1

Answer:

The density of solution is 1.283 g/mL.

Explanation:

Molarity of the KOH before dilution = $M_1$

Volume of the solution before dilution = $V_1=30.2 mL$

Molarity of the KOH after dilution = $M_2=0.173 M$

Volume of the solution after dilution = $V_2=1.20 L=1200 mL$

$M_1V_1=M_2V_2$

$M_1=\frac{M_2V_2}{V_1}=\frac{0.173 M\times 1200 mL}{30.2 mL}$

$M_1=6.8742 M$

$Molarity=\frac{moles}{Volume (L)}$

$V_1=30.2 mL=0.0302 L$ (1 mL = 0.001 L)

$M_1=\frac{n}{V_1}$

$n=M_1\times V_1=6.8742 M\times 0.0302 L=0.2076 mol$

Mass of 0.2076 moles of KOH:

0.2076 mol × 56 g/mol = 11.6256 g

Mass of KOH is solution = 11.6265 g

Mass of the solution = M

Mass percentage of solution = 30.0% of KOH

$30.0\%=\frac{11.6265 g}{M}\times 100$

M = 38.755 g

Density of the solution , d= $\frac{M}{V_1}$

$d=\frac{38.755 g}{30.2 mL}=1.283 g/mL$

The density of solution is 1.283 g/mL.