An aqueous solution of #H_2SO_4# (molecular weight=98) contains 10.78 #g# of acid per #dm^3# of solution. Density of solution is 1.123 #g#/#ml#. What is the molarity, molality, normality and mole fraction of the solution?

Question

Olivia Anton · Accepted Answer

Here's how you can solve this problem.

The problem actually makes your life easier by telling you that you have that much sulfuric acid, 10.78 g to be precise, per #"dm"^3# of solution. If you take into account that fact that #"1 dm"^3 = "1 L"#, you'll notice that all you need to do in order to determine the solution's molarity is calculate how many moles of sulfuric acid you have in 10.78 g. Use the molecular mass of the acid to do that. #10.78cancel("g") * ("1 mole"H_2SO_4)/(98cancel("g")) = "0.110 moles"# #H_2SO_4# This indicates that the molarity of the solution is #C = n_(H_2SO_4)/V = "0.110 moles"/"1 L" = color(green)("0.11 M")# Molality can be defined as the mass of the solvent (in kilograms) divided by the moles of solute (which you have already calculated)! Use the density of the solution to find the total weight of the solution in order to calculate the mass of water you have. #1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.123 g"/(1cancel("mL")) = "1123 g"# This implies that the water's mass will be #m_"water" = m_"solution" - m_(H_2SO_4)# #m_"water" = 1123 - 10.78 = "1112.2 g"# Thus, the solution's molality will be #b = n_(H_2SO_4)/m_"water" = "0.110 moles"/(1112.2 * 10^(-3)"kg") = color(green)("0.099 molal")# The amount of protons that sulfuric acid will produce in solution will be taken into account when determining the normality of the solution. Since sulfuric acid is a diprotic acid, every mole of the acid will produce two moles of #H^(+)# in solution, which are called equivalents. You will receive twice as many equivalents because you have 0.11 moles. Accordingly, the normality of the solution will be twice as large as its molarity. #N = ("equivalents of H"^(+))/"liter of solution" = "0.22 equiv."/("1 L") = color(green)("0.22 N")# You must first ascertain how many moles of water you have in order to obtain the mole fraction of the solution. #1112.2cancel("g") * "1 mole"/(18.02 cancel("g")) = "61.7 moles"# #H_2O# There will be a total of moles of #n_"total" = n_"water" + n_(H_2SO_4)# #n_"total" = 61.7 + 0.11 = "61.81 moles"# Sulfuric acid's mole fraction will therefore be #chi_(H_2SO_4) = n_(H_2SO_4)/n_"total" = (0.110cancel("moles"))/(61.81cancel("moles")) = color(green)("0.0018")#

Naomi Aronson · Answer

undefined Molarity: 1.1 M
Molality: 1.08 m
Normality: 2.2 N
Mole fraction: 0.04