# Question d8f3a

Jan 12, 2015

The answer is $\text{0.17 L}$, or $\text{170 mL}$.

You know you are dealing with a $\text{98%}$ $\text{v/v}$ solution of ${H}_{2} S {O}_{4}$. Since a $\text{v/v}$ percent concentration expresses the volume of the solute divided by the total volume of the solution and multiplied by 100%, your initial $\text{98%}$ $\text{v/v}$ solutin will have, for every 100.0 mL, 98.0 mL of ${H}_{2} S {O}_{4}$. So,

"98% v/v" = ("98.0 mL sulfuric acid")/("100 mL solution")

SInce we know that the solution's density is $\text{1.80 g/mL}$, we can determine what mass of ${H}_{2} S {O}_{4}$ we have for every 100 mL of solution

$\rho = \frac{m}{V} \implies {m}_{{H}_{2} S {O}_{4}} = \rho \cdot V = 1.80 \frac{g}{m L} \cdot 98.0 m L$

${m}_{{H}_{2} S {O}_{4}} = 176.4$ $\text{g}$

The goal is to determine what the molarity of the concentrated sulfuric acid, and then use this value to perform a simple dilution calculation in order to calculate the volume needed.

Since molarity is expressed in moles of solute per volume of solution, you need the moles of ${H}_{2} S {O}_{4}$, which come from

${n}_{{H}_{2} S {O}_{4}} = \frac{m}{\text{molar mass") = ("176.4 g")/("98.0 g/mol}} = 1.80$ $\text{moles}$

This means that the molarity of the concentrated solution is

$C = \frac{n}{V} = \left(\text{1.80 moles")/("100" * 10^(-3) "L}\right) = 18$ $\text{M}$

The volume of concentrated solution needed is

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

V_1 = ("0.5 M")/("18 M") * "6.00 L" = "0.17 L"#, or ${V}_{1} = \text{170 mL}$