Question

Calculate the mole fraction and molality of $HN{O_3}$, in a solution containing 12.2% $HN{O_3}$.[Given- atomic masses: H = 1, N = 14, O = 16]

Answer 1

To calculate the mole fraction of nitric acid, divide the moles of nitric acid with the total moles of both solute and solvent. To find out the molality of the solution divide the number of moles of nitric acid with the mass of solution in Kg.

Complete step by step answer:
Given,
Percentage of $HN{O_3}$ solution is 12.2 %
Atomic mass of hydrogen is 1.
Atomic mass of nitrogen is 14.
Atomic mass of oxygen is 16.
12.2 % $HN{O_3}$solution means, 12.2 g of $HN{O_3}$ is present in 100 g solution.
The molar mass of $HN{O_3}$is 63 g/mol.
The formula for calculating the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles.
m is the given mass
M is the molar mass
To calculate the number of moles of nitric acid ($HN{O_3}$), substitute the values of mass and molar mass in the above equation.
$\Rightarrow n = \dfrac{{12.2g}}{{63g/mol}}$
$\Rightarrow n = 0.194mol$
The mass of water is 87.8 g.
The molar mass of water is 18g/mol.
To calculate the number of moles of water substitute the values of mass and molar mass in the above equation.
$\Rightarrow n = \dfrac{{87.8g}}{{18g/mol}}$
$\Rightarrow n = 4.88mol$
The formula to calculate the mole fraction is shown below.
${X_A} = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}$
Where,
${X_A}$ is the mole fraction of compound A.
${n_A}$ is the number of moles of compound A.
${n_B}$ is the number of moles of compound B.
To calculate the mole fraction of $HN{O_3}$, substitute the values in the above equation.
$\Rightarrow {X_{HN{O_3}}} = \dfrac{{0.194mol}}{{0.194mol + 4.88mol}}$
$\Rightarrow {X_{HN{O_3}}} = 0.0396$
The molality of the solution is defined as the number of solute dissolved in 1kg of solution.
The formula to calculate molality is shown below.
$m = \dfrac{n}{{Kg\;of\;solution}}$
Where,
m is the molality.
n is the number of moles.
1000 g solution = 1 kg.
To calculate the molality, substitute the value of moles and mass of solution in kg in the above equation.
$\Rightarrow m = \dfrac{{0.194mol}}{{1kg}}$
$\Rightarrow m = 0.194m$
Therefore, the mole fraction of $HN{O_3}$ is 0.0396 and the molality is 0.194 m.

Note:
The percentage of the solution is present in mass by mass percentage, where the percentage given is the mass of solute and the remaining percent in the mass of solvent which forms a 100 % solution.