Question

Concentrated aqueous sulphuric acid is 98% ${H_2}S{O_4}$ by mass and has a density of 1.80 g $m{L^{ - 1}}$. The volume of acid required to make 1 litre of 0.1 M ${H_2}S{O_4}$ solution is:
A. 5.55 mL
B. 11.10 mL
C. 33.00 mL
D. 22.2O mL

Answer 1

In this question two molar solutions are prepared by the process of dilution. 98% mass of solution contains 98 g of sulphuric acid in 100 g solution. The volume of the first solution is calculated by the formula of density.

Complete step by step answer:
Given,
The percentage of ${H_2}S{O_4}$ is 98% by mass.
The density is 1.80 g $m{L^{ - 1}}$
The volume of acid is 1 litre
The molarity of ${H_2}S{O_4}$ solution is 0.1 M.
The percentage of ${H_2}S{O_4}$ is 98% by mass means 98 g of ${H_2}S{O_4}$ is present in 100 g solution.
The formula of density is shown below.
$D = \dfrac{M}{V}$
Where,
D is the density
M is the mass
V is the volume
To calculate the volume of solution, substitute the values in the above equation.
$\Rightarrow 1.80g/mL = \dfrac{{100g}}{V}$
$\Rightarrow V = \dfrac{{100g}}{{1.80g/mL}}$
$\Rightarrow V = 55.55mL$
The formula to calculate the number of moles is given below.
$n = \dfrac{m}{M}$
n is number of moles.
m is given mass.
M is molecular weight

The molecular weight of sulphuric acid is 98 g/mol.
To calculate the moles of sulphuric acid, substitute the values in the above equation.
$\Rightarrow n = \dfrac{{98g}}{{98g/mol}}$
$\Rightarrow n = 1mol$
The formula for calculating molarity is shown below.
$M = \dfrac{n}{V}$
Where,
M is the molarity
n is the number of moles
V is the volume.
To calculate the molarity of the solution, substitute the values in the above equation.
$\Rightarrow M = \dfrac{{1mol}}{{0.0555L}}$
$\Rightarrow M = 18.01M$
To calculate the volume of acid required to make 1 litre of 0.1 M ${H_2}S{O_4}$ solution, the formula of dilution is used.
${M_1} \times {V_1} = {M_2} \times {V_2}$
Where,
${M_1}$ is the molarity of solution 1.
${M_2}$ is the molarity of solution 2.
${V_1}$ is the volume of solution 1.
${V_2}$ is the volume of solution 2.
To find out the volume of sulphuric acid solution substitute the values in the above equation.
$\Rightarrow 18.02M \times {V_1} = 0.1M \times 1L$
$\Rightarrow {V_1} = \dfrac{{0.1M \times 1L}}{{18.02M}}$
${V_1} = 0.0055L$
Thus, the volume of acid required to make 1 litre of 0.1 M ${H_2}S{O_4}$ solution is 5.55 mL.

Therefore, the correct option is A.

Note: Make sure to convert the volume in millilitre into litres as the molarity is calculated in terms of moles per litre. Dilution is the process by which the molarity of higher concentration is converted into lower concentration.