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#
