Question

"what is the mole fraction, x, of solute and the molality, m (or b), for an aqueous solution that is 16.0% naoh by mass?"

Answer 1

Answer: Mole Fraction of solute = 0.0790

                Molality of solute = 4.761 mol/kg

Explanation: We are given that NaOH is 16% by mass in an aqueous solution which means that in 100gms of solution there is 16gm of NaOH present.

Mass of NaOH in aqueous solution = 16gms

Mass of water in aqueous solutin = 84gms

We need to calculate the mole fraction and molality of the solute and here solute is NaOH because solute is that substance in the solution which is present in low quantity.

To calculate the mole fraction, we will use the formula

$\chi =\frac{\text{moles of solute}}{\text{total moles in a solution}}$

and now to calculate mole fraction, we first need to find out the moles which can be calculated by

$\text{number of moles} =\frac{\text{Given mass}}{\text{molar mass}}$

Molar mass of NaOH = 40g/mol

Molar mass of $H_2O$ = 18g/mol

$\text{Moles of NaOH} =\frac{\text{16g}}{\text{40g/mol}}$

Moles of NaOH = 0.4mol

$\text{Moles of water}=\frac{\text{84g}}{\text{18g/mol}}$

Moles of water = 4.66mol

So, $\text{Mole fraction of NaOH}=\frac{\text{0.4mol}}{\text{0.4mol+4.66mol}}$

Mole fraction of NaOH = 0.0790

Now, to calculate molality, we use the formula

$\text{molality}=\frac{\text{moles of solute}}{\text{mass of solvent(in kg))}}$

mass of solvent (in kg) = $\frac{84g}{1000g/kg}$

Mass of solvent = 0.084kg

and we have already calculated the moles of solute above, by putting the values in the above equation, we find the molality.

$\text{molality of NaOH}= \frac{0.4}{0.084kg}$

Molality of NaOH = 4.761 mol/kg.