# The molality of a 1L solution with x% H2SO4 is 9. The weight of solvent is 910 grams. The value of x is?

Jul 31, 2017

#### Answer:

We have approx. 50% ${H}_{2} S {O}_{4}$ $\frac{w}{w}$........

#### Explanation:

We are given a concentration in terms of molality.

$\text{molality"="Moles of solute"/"Kilograms of solute} = 9 \cdot m o l \cdot k {g}^{-} 1$.

And thus....................................

$\text{Moles of solute} = 9 \cdot m o l \cdot k {g}^{-} 1 \times 0.910 \cdot k g = 8.19 \cdot m o l$, the which represents $8.19 \cdot m o l \times 98.08 \cdot g \cdot m o {l}^{-} 1 = 803.28 \cdot g$ with respect to sulfuric acid.

And thus...........

%"sulfuric acid m/m"="mass of sulfuric acid"/"mass of solution"=(803.28*g)/(803.28*g+910*g)xx100%

=46.9%=D as required.......

Do you see what I have done here? We were given the molality, and then we used this value to calculate the percentage by mass. The total mass included the sulfuric acid and the water...Quite clearly, we also had to KNOW definitions of $\text{molality}$, $m o l \cdot k {g}^{-} 1$, and also percentage by mass......"mass of solute"/"mass of solute + mass of solvent"xx100%.