What is the mass percent, molality and mole fraction of a solution made by dissolving 4g of sodium hydroxide in 90 g of water?

1 Answer
Jan 4, 2017

Answer:

#"Mass percent"# #=# #(4*g)/(90*g+4*g)xx100%=4.3%#

Explanation:

#"Molality"# #=# #"Moles of solute"/"Kilograms of solvent"#

#=((4*g)/(40.0*g*mol^-1))/(90xx10^-3*kg)=??*mol*kg^-1#

#"Mole fraction"="Moles of component"/"Sum of moles of all components of solution"#

#((4*g)/(40.0*g*mol^-1))/((4*g)/(40.0*g*mol^-1)+(90*g)/(18.01*g*mol^-1))=0.0196#.

For the purposes of illustration, we now calculate the mole fraction of water in the solution:

#((90*g)/(18.01*g*mol^-1))/((4*g)/(40.0*g*mol^-1)+(90*g)/(18.01*g*mol^-1))=0.979#.

The sum of the mole fractions is unity, as required.

Given that the molality will be near as dammit identical to molarity in suh a solution, what is the #pH# of this solution?