Question

Calculate molarity and molality of $6.3\%$ solution of nitric acid having density $1.04gc{{m}^{-3}}$(H=1, N=14, O=16)

Answer 1

The calculation of molarity and molality which is dependent on the definition of the terms and is given by the formula, Molarity =$M=\dfrac{{{m}_{solute}}}{{{m}_{solution(litres)}}}$ and molality =$m=\dfrac{{{m}_{solute}}}{{{m}_{solvent(kg)}}}$

Complete step by step solution:
We have studied about the basic concepts in the physical chemistry part about the molarity,molality and mole fraction of the solute and also about the normality of the solution.
Now, let us discuss the definitions of molarity and molality that helps us to deduce the correct answer.
- Molarity is defined as the number of moles of a solute present in one litre of the solution and thus the formula is given by,
$M=\dfrac{{{m}_{solute}}}{{{m}_{solution(litres)}}}$
- On the other hand molality is defined as the number of moles of solute that is present in 1 kg of the solvent and thus the formula of molality is given by,
$m=\dfrac{{{m}_{solute}}}{{{m}_{solvent(kg)}}}$
ow, density of the nitric acid solution according to the data given is$1.04gc{{m}^{-3}}$
Also, the volume of 100 grams of this solution will be = $\dfrac{100}{1.04}=96.15c{{m}^{3}}$
The above factor can also be converted into$d{{m}^{3}}$ as $96.15\times {{10}^{-3}}d{{m}^{3}}$
Therefore, the molar mass of the solute that is nitric acid is = $\left( 1\times 1 \right)+\left( 14\times 1 \right)+\left( 16\times 3 \right)=63g/mol$
Therefore, molarity of the solution will be,
$M=\dfrac{{{m}_{solute}}}{{{M}_{solute}}\times {{V}_{solution(litres)}}}$
where, ${{M}_{solute}}$ is the molar mass of the solute.
By, substituting the values, we have
$M=\dfrac{6.3}{63\times 96.15\times {{10}^{-3}}}=1.04mol/L$
Thus, molarity of the solution is$1.04mol/L$
Now, $6.3$ $HN{{O}_{3}}$ means that 6.3 g of nitric acid is present in 100 g of solution.
Therefore, mass of water will be =$100-6.3\text{ }=\text{ }93.7\text{ }g$ =$93.7\times {{10}^{-3}}kg$
Thus, molality of the solution will be,
$m=\dfrac{{{m}_{solute}}}{{{M}_{solute}}\times {{m}_{solvent(kg)}}}$
Thus, by substituting the values,
$m=\dfrac{6.3}{63\times 93.7\times {{10}^{-3}}}$
$\Rightarrow m=1.067mol/kg$

Therefore, molality of the solution will be $m=1.067mol/kg$.

Note: Do not be confused between the formulas of molarity, molality and normality of the solution as molarity is the measure of number of moles of solute per litre of solution, molality is the measure of number of moles of solute per kg of the solvent and normality is the measure of number equivalents per litre of the solution.