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"#?

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"#?

Wyatt Adkins · Accepted Answer

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

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#

Cooper Anderson · Answer

undefined 1. 2.77 mol/dm³
2. 0.18 dm³