Question

(i) Concentrated nitric acid used as a laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contained 23g of HNO3. Density of conc. HNO3 solution is 1.41g cm³.
(ii) The mole fraction of benzene in a solution of toluene is 0.40. Calculate the weight percent of benzene in the solution.

Answer 1

by this method you can do

Answer 2

Explanation:

(i)

23 gm of concentrated is present in $23.6 \mathrm{cm}^{3}$ volume of nitric acid.

Weight of concentrated $\mathrm{HNO}_{3}$= 100 gm

Weight of $\mathrm{HNO}_{3}$ = 69% = 69 gm

Weight of water = 31 gm

$d=\frac{m}{v}$

$v=\frac{m}{d}$

$v=\frac{100 \mathrm{gm}}{1.41 \frac{g}{\mathrm{cm}^{3}}}$

$=70.92 \mathrm{cm}^{3}$

Hence, volume of 69 gm of concentrated $\mathrm{HNO}_{3}$is 70.92 cm3.

23 gm of concentrated $\mathrm{HNO}_{3}$ is

$V=\frac{23 \times 70.92 \mathrm{cm}^{3}}{69} = 23.6 \mathrm{cm}^{3}$

Therefore, 23 gm of concentrated is present in $23.6 \mathrm{cm}^{3}$  volume of nitric acid.

(ii)

The weight percent of benzene in the solution is 44.82%.

Mole fraction of benzene = $X_{1}$

Mole fraction of toluene = $X_{2}$

For solution $X_{1}+X_{2}=1$

Molar mass of benzene = 78 g/mol

Molar mass of toluene = 96 gm

0.4 mol of benzene = 78 \times 0.4=31.2 gm

0.4 mol of toluene = 96 \times 0.4=38.4 gm

$\%=\frac{\text { Mass of component }}{\text {Mass of solution}} \times 100$

$\% \text { of benzene }=\frac{31.2}{31.2+38.4}=44.82 \%$