Question

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"`?

Answer 1

The concentration is `"9.1 mol/dm"^3`. You will need to use `"270 cm"^3` of the stock solution.

Explanation:

Molarity of `"Q"`:

Assume that you have `"1 dm"^3` of the stock solution.

`"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"`

`"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"`

`"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"`

`"Molarity" = "9.14 mol"/("1 dm"^3) = "9.1 mol/dm"^3`

Volume of stock solution

`c_1V_1 = c_2V_2`

`V_1 = V_2 × c_2/c_1`

In this problem,

`V_1 = "?"; color(white)(mmml)c_1 = "9.1 mol/dm"^3`
`V_2 = "500 cm"^3; c_2 = 5 color(white)(m)"mol/dm"^3`

`V_1 = "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`