Question

Calculate the mole fraction of NaOH in 10% w/w aqueous solution. [At. MassH=1, O=16, Na=23u]​

Answer 1

Answer:

0.047 is your answer.See the attachment.

Answer 2

Answer:

 The mole fraction of $NaOH$ measured is $0.047$.

Explanation:

Given,

The $10\%$ w/w aqueous solution of $NaOH$.

The atomic mass of hydrogen = $1u$

The atomic mass of oxygen = $16u$

The atomic mass of sodium = $23u$

The mole fraction of $NaOH$ in $10\%$ w/w aqueous solution =?

As given,

• $10\%$ w/w aqueous solution of $NaOH$.
• This means, $10g$ of $NaOH$ present in $100g$ of the solution.

Therefore,

• Mass of solvent ( $H_{2}O$ ) = Mass of the solution - the mass of $NaOH$
• Mass of solvent ( $H_{2}O$ ) = $100-10$ = $90g$

Now,

• The molar mass of $NaOH$ = atomic mass of Na + atomic mass of O + atomic mass of H
• The molar mass of $NaOH$ = $23 + 16 + 1$ = $40u$ = $40g/L$

Also,

• The molar mass of $H_{2}O$ = 2 × atomic mass of H + atomic mass of O
• The molar mass of $H_{2}O$ = $2\times1 + 16$ = $18u$ = $18g/L$.

Now,

• The number of moles of $NaOH$ = $\frac{Given mass }{Molar mass}$ = $\frac{10}{40}$ = $0.25 mol$
• The number of moles of $H_{2}O$ = $\frac{Given mass }{Molar mass}$ = $\frac{90}{18}$ = $5 mol$

Therefore,

• The mole fraction of $NaOH$ = $\frac{n_{NaOH} }{n_{NaOH} + n_{H_{2}O } }$ = $\frac{0.25}{0.25 +5}$ = $0.047$.