Question

In what ratio should a 90 per cent solution of ${H_2}S{O_4}$ be mixed with a 10 per cent solution of the acid to obtain a 40 per cent solution?
A) 5:3
B) 3:4
C) 3:5
D) 4:5

Answer 1

Hint: We have to calculate the ratio of using the molarity equation.
Formula used: Volumetric titrations could be calculated using the formula,
${M_2}{V_2} = {M_1}{V_1}$
Where ${{\text{M}}_{\text{1}}}$ is the molarity of the titrant.
${{\text{M}}_{\text{2}}}$ is the molarity of the analyte.
${{\text{V}}_{\text{1}}}$is the volume of the titrant.
${{\text{V}}_2}$ is the volume of the analyte.

Complete step by step answer:
We know that the density of a substance is the mass to the volume. We can write the formula as,
${\text{Density}} = \dfrac{{{\text{Mass}}\left( g \right)}}{{{\text{Volume}}\left( {c{m^3}} \right)}}$
We can rearrange the equation for the mass as,
\[{\text{Mass}} = {\text{Volume}} \times {\text{Density}}\]
Here, Density is constant and therefore, the mass is proportional to volume.
The volumetric titration formula is,
${M_2}{V_2} = {M_1}{V_1}$
Let ${V_1}$ mL be 90% and ${V_2}$ mL of 10% ${H_2}S{O_4}$ be mixed. Using the molarity equation,
Mass percent of the obtained solution is 40%.
Let us substitute the values of mass percent to calculate the ratio.
${M_1}{V_1} + {M_2}{V_2} = M\left( {{V_1} + {V_2}} \right)$
$90{V_1} + 10{V_2} = 40\left( {{V_1} + {V_2}} \right)$
$90{V_1} + 10{V_2} = 40{V_1} + 40{V_2}$
$50{V_1} = 30{V_2}$
Let us now divide ${V_1}$ mL by ${V_2}$ mL to get the ratio
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{30mL}}{{50mL}}$
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{3mL}}{{5mL}}$
So the ratio of ${V_1}:{V_2} = 3:5$.

So, the correct answer is Option C.

Note:
We can calculate the mass percent using the formula,
${\text{Mass percentage}} = \dfrac{{{\text{Mass}}}}{{{\text{Total mass}}}} \times 100\% $
Example: Given,
Molarity of the solution is $2.77M$
Density of the solution is $1.109g/ml$
The grams of the sodium hydroxide are calculated using the molar mass.
Grams of sodium hydroxide = $2.77mol \times \dfrac{{39.97g}}{{1mol}} = 110.71g$
The mass of the solution is calculated using the density.
Mass of the solution = $1L \times \dfrac{{1.109g}}{{1ml}} \times \dfrac{{1000ml}}{{1L}} = 1109g$
The concentration of the solution is,
${\text{Mass percentage}} = \dfrac{{{\text{Grams of solute}}}}{{{\text{Grams of solution}}}} \times 100\% $
Concentration of the solution = $\dfrac{{110.71g}}{{1109g}} \times 100\% $
Concentration of the solution = $9.982\% $
The concentration of the solution expressed in mass percentage is ${\text{9}}{\text{.982\% }}$.