# How do you find the volume of concentrated 18M sulfuric acid to prepare 250mL of a 6.0M solution?

Mar 4, 2017

We use the relationship: $\text{concentration"="Moles of solute"/"Volume of solution}$, and get under $100 \cdot m L$ of conc. acid.

#### Explanation:

We require $250 \cdot m L$ of a $6.0 \cdot m o l \cdot {L}^{-} 1$ solution (with respect to H_2SO_4).

With respect to ${H}_{2} S {O}_{4}$, this represents a molar quantity of $0.250 \cdot L \times 6.0 \cdot m o l \cdot {L}^{-} 1 = 1.50 \cdot m o l$.

If the mother solution has a $18 \cdot m o l \cdot {L}^{-} 1$ concentration, then we solve the quotient, (1.50*cancel(mol))/(18*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)=??*mL.

Alternatively we could use the following formula for the molar equivalence:

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$.............

$\text{Important, so get it right.................}$

$\text{WHEN YOU DO THESE DILUTIONS YOU ALWAYS ADD}$
$\text{ACID TO WATER!!!!}$

$\text{Why?}$ BECAUSE IF YOU SPIT IN ACID IT SPITS BACK AT YOU. I am not joking.