# Question a6138

Dec 9, 2014

The answer is $2.0 \cdot {10}^{2} \text{mL}$.

A solution's percent concentration by mass is defined as the mass of the solute, in your case hydrochloric acid, divided by the total mass of the solution, and multiplied by 100.

You know that the stock solution has a concentration of 37%, which means that you get 37 g of hydrochloric acid for every 100 g of solution.

Now, let's assume we have a 1-L bottle of this solution (you can assume any volume for the bottle, the answer is the same - 1L just made the calculations easier). Its density is $\text{1.20 g/mL}$, which is equivalent to

$1.2 \text{g"/cancel("mL") * (1000 cancel("mL"))/"1 L" = "1200 g/L}$.

You can use the solution's density to determine the mass of the 1-L solution

1cancel("L") * "1200 g"/(1cancel("L")) = "1200 g"

Now, you know that for every 100 g of solution you have 37 g of hydrochoric acid. This means that you get

1200cancel("g solution") * "37 g HCl"/(100cancel("g solution")) = "444 g HCl"

Now use hydrochloric acid's molar mass to determine how many moles you get in that much solution

444cancel("g") * "1 mole HCl"/(36.46cancel("g")) = "12.18 moles HCl"

The stock solution's molarity will thus be

C = n/V = ("12.18 moles")/("1 L") = "12.2 M"

Use the dilution concentration formula, to get the volume of the stock solution needed to make your target solution

${C}_{1} {V}_{1} = {C}_{2} {V}_{2} \implies {V}_{1} = {C}_{2} / {C}_{1} \cdot {V}_{2}$

V_1 = (0.700cancel("M"))/(12.2cancel("M")) * 3.5 * 10^3"mL" = "200.8 mL"#

Rounded to two sig figs, the number of sig figs you gave for the volume of the target solution, the answer will be

${V}_{1} = \textcolor{g r e e n}{2.0 \cdot {10}^{2} \text{mL}}$