# Question cd3b0

May 24, 2017
1. $12 \text{mol"/"L}$

2. $0.081 \text{L" = 81"mL}$

#### Explanation:

1.

Let's look at the problem: the $\text{HCl}$ in solution is 38% by mass. If we assume a $100 \text{g}$ sample, then then the mass of $\text{HCl}$ in solution is $38 \text{g}$. The number of moles of the solute is thus

38cancel("g HCl")((1 "mol HCl")/(36.46cancel( "g HCl"))) = 1.04 "mol HCl"

Now, let's use the density of the solution ($1.19 \text{g"/"mL}$), and the fact that we assumed a $100 \text{g}$ sample to calculate the volume, in $\text{L}$:

100 cancel("g")((1cancel( "mL"))/(1.19 cancel("g")))((1 "L")/(10^3 cancel("mL"))) = 0.0840 "L soln"

The molarity of the solution is thus

$M = \text{mol solute"/"L soln" = (1.04 "mol HCl")/(0.0840 "L soln") = color(red)(12"mol"/"L}$

2.

In $1000 \text{mL}$ (or $1 \text{L}$) of a $1 M$ solution, the number of moles of $\text{HCl}$ is

$1 \text{mol"/cancel("L")(1 cancel("L")) = 1 "mol HCl}$

We'll use this number and the fact that $1.04 \text{mol HCl}$ occupies a volume of $0.0840 \text{L}$ to find the volume of solution needed:

$1 \cancel{\text{mol HCl")((0.0840 "L soln")/(1.04 cancel("mol HCl"))) = color(blue)(0.081 "L}}$, or color(blue)(81 "mL"#