Question

Concentrated $HN{O_3}$, is 63% by weight (density = 1.4 gm/mL). The volume of this acid required to prepare 250 mL of 1.2 M solution is:
A. 12
B. 15.7
C. 21.4
D. 42.6

Answer 1

Hint: The concentration of the solution is defined as the number of moles of solute present in one litre of solution. To calculate the volume of acid, first the molarity of the acid is needed to be determined.

Complete step by step answer:
Given,
Concentration of $HN{O_3}$ is 63%.
Density is 1.4 g/mL.
Volume of solution is 250 mL
Molarity of the solution is 1.2 M.
63% weight means 63 gm of $HN{O_3}$ is present in 100 g of solution.
The formula for calculating the moles is shown below.
$n = \dfrac{m}{M}......(i)$
Where
n is the number of moles of compound.
m is the mass of the compound.
M is the molar mass of the compound.
To calculate the moles of $HN{O_3}$, substitute the value of mass and atomic mass in equation (i).
$n = \dfrac{{63g}}{{63g/mol}}$
$\Rightarrow n = 1mol$
The volume of the solution is calculated by the formula shown below.
$V = \dfrac{m}{d}$
Where,
V is the volume
m is the mass
d is the density
To calculate the volume of solution, substitute the value of mass and density in the above equation.
$V = \dfrac{{100g}}{{1.4g/mL}}$
$\Rightarrow V = 71.428mL$
The formula for calculating the molarity is shown below.
$M = \dfrac{n}{V}$
M is the molarity
n is the number of moles.
V is the volume.
To calculate the molarity of $HN{O_3}$, substitute the values in the above equation.
$M = \dfrac{{1mol}}{{0.0714L}}$
$\Rightarrow M = 14M$
The calculation for the volume of $HN{O_3}$ solution required to form 1.2 M of 250 mL is shown below.
${M_1}{V_1} = {M_2}{V_2}$
$\Rightarrow {V_1} = \dfrac{{{M_2}{V_2}}}{{{M_1}}}$
$\Rightarrow V = \dfrac{{1.2M \times 250mL}}{{14M}}$
$\Rightarrow {V_1} = 21.4mL$

Therefore, the correct option is C .

Note: This method is used in qualitative volumetric analysis for the dilution process. The dilution is defined as the change in concentration of solution by adding extra solvent without adding the solute.