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
Nov 28, 2016

Answer:

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#