In this order, rho = 1.8g/(mL), 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 100g of solution, 98g represent sulphuric acid, H_2SO_4.
We know that 1dm^3 = 1L, so H_2SO_4's molarity is
C = n/V = (18.0 mol es)/(1.0L) = 18M
In order to determine sulphuric acid solution's density, we need to find its mass; H_2SO_4's molar mass is 98.0g/(mol), so
(18.0 mol es)/(1L) * (98.0g)/(1 mol e) = 1764g/(1L)
Since we've determined that we have 1764g of H_2SO_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)) ->
mass_(solution) = (1764g * 100.0g)/(98g) = 1800g
Therefore, 1L of 98 wt% H_2SO_4 solution will have a density of
rho = m/V = (1800g)/(1.0 * 10^3 mL) = 1.8g/(mL)
H_2SO_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_2SO_4)/(mass_(solvent)) = (18 mol es)/((1800-1764) * 10^(-3)kg) = 0.5m
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 18g/(mol), we could determine that
100g.solution * (98g)/(100g) * (1 mol e)/(98g) = 1 mole H_2SO_4
100g.solution * ((100-98)g)/(100g) * (1 mol e)/(18g) = 0.11 moles H_2O
So, H_2SO_4's mole fraction is
mol ef r a ction_(H_2SO_4) = 1/(1+0.11) = 0.9
