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"?

1 Answer
Ernest Z.
Nov 28, 2016

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

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