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

Jan 4, 2017

$\text{Mass percent}$ $=$ (4*g)/(90*g+4*g)xx100%=4.3%

#### Explanation:

$\text{Molality}$ $=$ $\text{Moles of solute"/"Kilograms of solvent}$

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

$\text{Mole fraction"="Moles of component"/"Sum of moles of all components of solution}$

$\frac{\frac{4 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1}}{\frac{4 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1} + \frac{90 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.0196$.

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

$\frac{\frac{90 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}}{\frac{4 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1} + \frac{90 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}} = 0.979$.

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 $p H$ of this solution?