# Given 93.2% "w/w" "sulfuric acid" that has rho=1.835*g*mL^-1, how should we dilute this to get a 3*N solution of 0.50*L volume...?

Jul 29, 2017

A messy problem.......we need a $43 \cdot m L$ volume of conc. acid which we add to water to make a $0.5 \cdot L$ volume.

#### Explanation:

First off, to settle the insurance issue, when you MAKE dilutions with acid and water YOU ALWAYS ADD ACID to water, and never the reverse. Why not? Because if you spit in acid it spits back; I kid you not.

We (i) determine the molarity of our sulfuric acid, and we want the quotient, $\text{moles of acid"/"volume of solution}$.

And thus $1 \cdot m L$ of the acid...........(I think there is a typo in your question)..............expresses a concentration of.....

((1*mLxx1.835*g*mL^-1xx93.2%)/(98.08*g*mol^-1))/(1xx10^-3*L)=17.4*mol*L^-1

Can you see what I have done with the quotient? The numerator represents the moles of sulfuric acid; the denominator represents the volume of the solution; I get a concentration as required.

Now we wants a $0.5 \cdot L$ volume of $3 N$ acid.......But sulfuric acid is a diacid, so we wants a $0.5 \cdot L$ volume of $1.5 \cdot m o l \cdot {L}^{-} 1$ ${H}_{2} S {O}_{4}$.

So we require, $1.50 \cdot m o l \cdot {L}^{-} 1 \times 0.5 \cdot L \equiv 0.75 \cdot m o l$ ${H}_{2} S {O}_{4}$.

And using the conc. acid we need ..............

$\frac{0.75 \cdot m o l}{17.5 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 \cong 43 \cdot m L$....