Question

The volume of ${{H}_{2}}S{{O}_{4}}$ acid (98% by mass, d = 1.80g/mL) required to prepare 1 litre of 0.1 M ${{H}_{2}}S{{O}_{4}}$ solution is:
(A) 16.65mL
(B) 22.20mL
(C) 5.55mL
(D) 11.10mL

Answer 1

As we know that molarity of a solution is the amount of a substance or solute present in 1 litre of solution. Here we have to calculate the volume of acid used to prepare its diluted solution, so we will use the concept of dilution along with other concentration terms involved.
Formula used:
We will use the following formulas:-
Molarity = $\dfrac{\text{number of moles of solute}}{\text{volume of solution (in L)}}$
For dilution:-
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$

Complete answer:
Let us first discuss about molarity and dilution as follows:-
Molarity: It is the amount of a substance or solute present in 1 litre of solution. Mathematically it can be represented as:-
Molarity = $\dfrac{\text{number of moles of solute}}{\text{volume of solution (in L)}}$
Dilution: It is a process of decreasing the concentration of a solute in a solution by increasing the concentration of solvent in it. The relation between the molarity and volume of before and afterward solution is${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$.
-Calculation of weight of ${{H}_{2}}S{{O}_{4}}$ acid and its solution as follows:-
Given that ${{H}_{2}}S{{O}_{4}}$ is 98% by mass which means 98 grams of ${{H}_{2}}S{{O}_{4}}$ acid is present in 100 grams of solution.
Mass of${{H}_{2}}S{{O}_{4}}$ acid = 98 grams
Mass of solution = 100 grams
-Calculation of volume of ${{H}_{2}}S{{O}_{4}}$ solution as follows:-
Density of solution = 1.80g/mL
Density = $\dfrac{\text{Mass of solution}}{\text{Volume of solution}}$
Rearrange the above formula for further calculations:-
$\begin{aligned} & \Rightarrow \text{Volume of solution}=\dfrac{\text{Mass of solution}}{\text{Density of solution}} \\\ & \Rightarrow \text{Volume of solution}=\dfrac{100g}{1.80g/mL} \\\ & \Rightarrow \text{Volume of solution}=55.55mL \\\ \end{aligned}$
Volume of solution in litres = 0.0555L
-Calculation of molarity (${{M}_{1}}$) of ${{H}_{2}}S{{O}_{4}}$solution before dilution i.e., concentrated ${{H}_{2}}S{{O}_{4}}$solution:-
Molar mass of ${{H}_{2}}S{{O}_{4}}$= 98g/mol
Number of moles of ${{H}_{2}}S{{O}_{4}}$= $\dfrac{\text{Mass of }{{\text{H}}_{2}}S{{O}_{4}}}{\text{Molar mass of }{{\text{H}}_{2}}S{{O}_{4}}}$
Number of moles of ${{H}_{2}}S{{O}_{4}}$=$\dfrac{98g}{98g/mol}=1mole$
Now let us use the formula of Molarity as follows:-
$\begin{aligned} & \Rightarrow {{M}_{1}}=\dfrac{\text{number of moles of solute}}{\text{volume of solution (in L)}} \\\ & \Rightarrow {{M}_{1}}=\dfrac{1mole}{0.0555L} \\\ & \Rightarrow {{M}_{1}}=18.02M \\\ \end{aligned}$-Now let us consider that ‘V’ mL of ${{H}_{2}}S{{O}_{4}}$are used in preparation of 1 litre (${{V}_{2}}$ ) of 0.1 M ${{H}_{2}}S{{O}_{4}}$solution.
${{V}_{2}}$in milliliters = 1000mL
So substitute all the values in the formula of dilution as follows:-
$\begin{aligned} & \Rightarrow {{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}} \\\ & \Rightarrow 18.02M\times V=0.1M\times 1000mL \\\ & \Rightarrow V=\dfrac{100M\times mL}{18.02M} \\\ & \Rightarrow V=5.55mL \\\ \end{aligned}$

Therefore volume of ${{H}_{2}}S{{O}_{4}}$ acid required to prepare 1 litre of 0.1 M ${{H}_{2}}S{{O}_{4}}$solution is: (C) 5.55mL.

Note:
-Always change the units of the values according to the given options or as per requirement so as to obtain the correct answer.
-Also understand the concept of dilution and laws of equivalence to solve such types of questions.