# What is molar concentration of Compound "Q" in a 31 % by mass stock solution with a density of "1.12 kg/dm"^3? What volume of the stock solution would you need to make "500 cm"^3 of "5 mol/dm"^3color(white)(l)"Q"?

Nov 28, 2016

The concentration is ${\text{9.1 mol/dm}}^{3}$. You will need to use ${\text{270 cm}}^{3}$ of the stock solution.

#### Explanation:

Molarity of $\text{Q}$:

Assume that you have ${\text{1 dm}}^{3}$ of the stock solution.

$\text{Mass of solution" = 1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"stock"))) × "1.12 kg stock"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"stock")))) = "1.12 kg stock" = "1120 g stock}$

$\text{Mass of Q" = 1120 color(red)(cancel(color(black)("g stock"))) × "31 g Q"/(100 color(red)(cancel(color(black)("g stock")))) = "347 g Q}$

$\text{Moles of Q" = 347 color(red)(cancel(color(black)("g Q"))) × "1 mol Q"/(38.0 color(red)(cancel(color(black)("g Q")))) = "9.14 mol Q}$

${\text{Molarity" = "9.14 mol"/("1 dm"^3) = "9.1 mol/dm}}^{3}$

Volume of stock solution

${c}_{1} {V}_{1} = {c}_{2} {V}_{2}$

V_1 = V_2 × c_2/c_1

In this problem,

${V}_{1} = {\text{?"; color(white)(mmml)c_1 = "9.1 mol/dm}}^{3}$
${V}_{2} = {\text{500 cm"^3; c_2 = 5 color(white)(m)"mol/dm}}^{3}$

${V}_{1} = {\text{500 cm"^3 × (5 color(red)(cancel(color(black)("mol/dm"^3))))/(9.1 color(red)(cancel(color(black)("mol/dm"^3)))) = "270 cm}}^{3}$