# Question 5998a

Aug 3, 2017

$\text{molarity} = 14.6$ $M$

$\text{volume} = 13.7$ $\text{mL}$

#### Explanation:

We're asked to find (a) the molarity of the ${\text{H"_2"SO}}_{4}$ solution, and (b) the volume of that acid solution required to make $1$ $\text{L}$ of a $0.2$ $M$ solution.

(a)

To find the molarity of the solution, we can first recognize that in an 80% by mass solution, and we assume a $100$-$\text{g}$ sample, there would be

• $80.0$ ${\text{g H"_2"SO}}_{4}$

• $20.0$ $\text{g H"_2"O}$

Also in a $100$-$\text{g}$ sample, using the given density, the volume (in liters) of the solution is

V = 100cancel("g soln")((1cancel("mL soln"))/(1.787cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(red)(ul(0.0560color(white)(l)"L soln"

Remember that the molarity equation is

ul("molarity" = "mol solute"/"L soln"

We just found the liters of solution, so let's use the $80$ ${\text{g H"_2"SO}}_{4}$ and the molar mass of ${\text{H"_2"SO}}_{4}$ to find the moles:

80.0cancel("g H"_2"SO"_4)((1color(white)(l)"mol H"_2"SO"_4)/(98.078cancel("g H"_2"SO"_4))) = color(green)(ul(0.816color(white)(l)"mol H"_2"SO"_4

The molarity of the solution is thus

"molarity" = color(green)(0.816color(white)(l)"mol H"_2"SO"_4)/(color(red)(0.0560color(white)(l)"L soln")) = color(blue)(ulbar(|stackrel(" ")(" "14.6color(white)(l)M" ")|)

(b)

Now, we can use the dilution equation to find the volume of this solution required to make $1$ $\text{L}$ of a $0.2$ $M$ solution.

The dilution equation is

ul(M_1V_1 = M_2V_2

where

• ${M}_{1}$ and ${M}_{2}$ are the molarities of the two solutions

• ${V}_{1}$ and ${V}_{2}$ are the respective volumes of the two solutions

We know:

• M_1 = color(blue)(14.6color(white)(l)M

• V_1 = ?

• ${M}_{2} = 0.2 \textcolor{w h i t e}{l} M$

• ${V}_{2} = 1 \textcolor{w h i t e}{l} \text{L}$

Let's rearrange the equation to solve for the unknown volume, ${V}_{1}$:

${V}_{1} = \frac{{M}_{2} {V}_{2}}{{M}_{1}}$

Plugging in known values:

V_1 = ((0.2cancel(M))(1color(white)(l)"L"))/(color(blue)(14.6)cancel(color(blue)(M))) = color(blue)(0.0137color(white)(l)"L" = color(blue)(ulbar(|stackrel(" ")(" "13.7color(white)(l)"mL"" ")|)#