Question

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

Answer 1

`"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?