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?

1 Answer
Jul 9, 2015

Answer:

Here's how you can solve this problem.

Explanation:

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.

To do that, use the acid's moalr mass

#10.78cancel("g") * ("1 mole"H_2SO_4)/(98cancel("g")) = "0.110 moles"# #H_2SO_4#

This means that the solution's molarity is

#C = n_(H_2SO_4)/V = "0.110 moles"/"1 L" = color(green)("0.11 M")#

Molality is defined as moles of solute, which you've already calculated, divided by the mass of the solvent - expressed in kilograms!

To get the mass of water you have, use the solution's density to first determine how much the entire solutions weighs.

#1cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.123 g"/(1cancel("mL")) = "1123 g"#

This means that the mass of water will be

#m_"water" = m_"solution" - m_(H_2SO_4)#

#m_"water" = 1123 - 10.78 = "1112.2 g"#

This molality of the solution will thus be

#b = n_(H_2SO_4)/m_"water" = "0.110 moles"/(1112.2 * 10^(-3)"kg") = color(green)("0.099 molal")#

The solution's normality will take into account the number of protons the sulfuric acid will produce in 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.

Since you have 0.11 moles, you'll get twice as many equivalents.

This means that the solution's normality will be twice as big as its molarity.

#N = ("equivalents of H"^(+))/"liter of solution" = "0.22 equiv."/("1 L") = color(green)("0.22 N")#

To get the mole fraction of the solution, you need to first determine how many moles of water you have.

#1112.2cancel("g") * "1 mole"/(18.02 cancel("g")) = "61.7 moles"# #H_2O#

The total number of moles will be

#n_"total" = n_"water" + n_(H_2SO_4)#

#n_"total" = 61.7 + 0.11 = "61.81 moles"#

The mole fraction of sulfuric acid will thus be

#chi_(H_2SO_4) = n_(H_2SO_4)/n_"total" = (0.110cancel("moles"))/(61.81cancel("moles")) = color(green)("0.0018")#