Calculate the molality of a sulfuric acid solution of specific gravity 1.2 containing 27% H_2SO_4 by **weight**?

Aug 7, 2017

$\left[{H}_{2} S {O}_{4}\right] \cong 4 \cdot m o l \cdot k {g}^{-} 1$

Explanation:

We want the expression......

$\text{molality"="moles of solute"/"kilograms of solvent}$.....

And our ${H}_{2} S {O}_{4} \left(a q\right)$ has a density of $\rho = 1.2 \cdot g \cdot m {L}^{-} 1$.

And thus we can work with a $1 \cdot m L$ $\text{volume..........}$

"Molality"=((1.2*gxx27%)/(98.08*g*mol^-1))/((1.2*g-0.324*g)xx10^-3*kg*g^-1)

$= 3.77 \cdot m o l \cdot k {g}^{-} 1$ with respect to ${H}_{2} S {O}_{4}$......

Here we multiplied that mass of solution by the percentage of acid, and then the percentage of water, to get the moles of solute, and moles of water solvent. Please check my figures...........

Just as an exercise, we could also interrogate the $\text{molarity}$ of the given solution, and here we want the quotient.....

$\text{Molarity"="Moles of solute"/"Volume of solution (L)}$....

And thus.....we calculate the concentration a ONE MILLILITRE of this acid solution......which has a mass of $1.20 \cdot g$....Agreed?

And so......"Molarity"=((1.2*gxx27%)/(98.08*g*mol^-1))/(1*mLxx10^-3*L*mL^-1)=3.30*mol*L^-1

.........for a more dilute solution, molality would often be identical to molarity. When we use sulfuric acid solution, the solution density is such that difference in the concentrations occur.