# Question e69de

Dec 17, 2014

In this order, $\rho = 1.8 \frac{g}{m L}$, ${c}_{m} = 0.5$, and mole fraction = $0.9$

First, let's start with wt%, which is the symbol for weight percent. 98 wt% means that for every $100 g$ of solution, $98 g$ represent sulphuric acid, ${H}_{2} S {O}_{4}$.

We know that $1 {\mathrm{dm}}^{3} = 1 L$, so ${H}_{2} S {O}_{4}$'s molarity is

$C = \frac{n}{V} = \frac{18.0 m o l e s}{1.0 L} = 18 M$

In order to determine sulphuric acid solution's density, we need to find its mass; ${H}_{2} S {O}_{4}$'s molar mass is $98.0 \frac{g}{m o l}$, so

$\frac{18.0 m o l e s}{1 L} \cdot \frac{98.0 g}{1 m o l e} = 1764 \frac{g}{1 L}$

Since we've determined that we have $1764 g$ of ${H}_{2} S {O}_{4}$ in 1L, we'll use the wt% to determine the mass of the solution

98.0 wt% = (98g. H_2SO_4)/(100.0g. solution) = (1764g)/(mass_(solution)) ->

$m a s {s}_{s o l u t i o n} = \frac{1764 g \cdot 100.0 g}{98 g} = 1800 g$

Therefore, $1 L$ of 98 wt%# ${H}_{2} S {O}_{4}$ solution will have a density of

$\rho = \frac{m}{V} = \frac{1800 g}{1.0 \cdot {10}^{3} m L} = 1.8 \frac{g}{m L}$

${H}_{2} S {O}_{4}$'s molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be

${c}_{m} = {n}_{{H}_{2} S {O}_{4}} / \left(m a s {s}_{s o l v e n t}\right) = \frac{18 m o l e s}{\left(1800 - 1764\right) \cdot {10}^{- 3} k g} = 0.5 m$

Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is $18 \frac{g}{m o l}$, we could determine that

$100 g . s o l u t i o n \cdot \frac{98 g}{100 g} \cdot \frac{1 m o l e}{98 g} = 1$ mole ${H}_{2} S {O}_{4}$

$100 g . s o l u t i o n \cdot \frac{\left(100 - 98\right) g}{100 g} \cdot \frac{1 m o l e}{18 g} = 0.11$ moles ${H}_{2} O$

So, ${H}_{2} S {O}_{4}$'s mole fraction is

$m o l e f r a c t i o {n}_{{H}_{2} S {O}_{4}} = \frac{1}{1 + 0.11} = 0.9$