Question

Concentrated aqueous sulphuric acid is 98% ${H_2}S{O_4}$ by mass and has a density of 1.80 g/mL. The volume of acid required to make 1 liter of 0.1 M ${H_2}S{O_4}$ solution is:
(A) 5.55 mL
(B) 11.10 mL
(C) 33.00 mL
(D) 22.20 mL

Answer 1

Hint: The formula to find the molarity of the solution is as below.
\[M = \dfrac{{{\text{Weight of solute}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{Volume of the solution(mL)}}}}\]

Complete Step-by-Step Solution:
From the given data, we will first find the molarity of sulphuric acid and then we will find the volume of 98% sulphuric acid solution required to prepare the given solution.
- We are provided that the solution of ${H_2}S{O_4}$ is 98% by mass. So, we can say that if the weight of solution is 100 g, then the weight of ${H_2}S{O_4}$ in that solution is 98 g. So, remaining 2 g will be water.
- We are given that the density of the solution of 98% ${H_2}S{O_4}$ is 1.80 g/mL.
So, if we assume a 100 mL solution of 98% ${H_2}S{O_4}$ , we can find the volume of the solution with the use of density. We know that
\[{\text{Density = }}\dfrac{{{\text{Weight}}}}{{{\text{Volume}}}}\]
So, $1.80 = \dfrac{{100}}{{{\text{Volume}}}}$
Thus, Volume = $\dfrac{{100}}{{1.80}} = 55.55mL$
Now, we know that the volume of the solution is 55.55mL. We also know that the weight of ${H_2}S{O_4}$ in the solution is 98 g.
Now, molecular weight of ${H_2}S{O_4}$ = 2(Atomic weight of H) + Atomic weight of S + 4(Atomic weight of O)
Molecular weight of ${H_2}S{O_4}$ = 2(1) + 32 + 4(16) = 98$gmo{l^{ - 1}}$
Now, we can find the molarity of 98% ${H_2}S{O_4}$ solution using the formula given below.
\[M = \dfrac{{{\text{Weight of solute}} \times {\text{1000}}}}{{{\text{Molecular weight of solute}} \times {\text{Volume of the solution(mL)}}}}\]
So, we can write that
\[M = \dfrac{{98 \times 1000}}{{98 \times 55.55}} = 18M\]
Thus, we obtained that the molarity of that solution is 18M.
Now, we can write that
\[{M_1}{V_1} = {M_2}{V_2}\]
Where ${M_1}$ = molarity of 98% ${H_2}S{O_4}$ solution
${V_1}$ = volume of 98% ${H_2}S{O_4}$ solution required
${M_2}$ = molarity of new solution = 0.1M
${V_2}$ = Volume of new solution = 1L
So, we obtain that
\[18 \times {V_1} = 0.1 \times 1\]
So, ${V_1} = \dfrac{{0.1 \times 1}}{{18}} = 0.005555$ L
Now, we obtained the volume in liters.
We know that 1 L = 1000mL
So, 0.0055 L = 1000 $ \times $ 0.0055 = 5.55 mL

Thus, we can say that the correct answer is (A).

Note: Remember that whenever we make a solution from another solution of known concentration, then we can use the formula ${M_1}{V_1} = {M_2}{V_2}$ to find the volume or molarity whichever is unknown. If we are given normality, then we can use normality of both the solutions in the same formula.