Question

A sample of ${{H}_{2}}S{{O}_{4}}$ (density 1.8 gram per ml) is 90% by weight. What is the volume of acid that has to be used to make 1 litre of 0.2 M ${{H}_{2}}S{{O}_{4}}$?
(A) 15
(B) 11.2
(C) 12.1
(D) 22.4

Answer 1

Hint: Molarity is the basic form of solving such types of problems. It is defined as the moles of solute present per litre of solution and denoted as M (moles of solute/litre of solution).

Complete answer:
Let us see the molarity in detail before actually solving the given illustration;
Molarity-
As we know, molarity is the moles of solute present per litre of solution i.e.
\[M=\dfrac{n}{V}\]
where,
M = molar concentration (molarity)
n = moles of solute
V = volume of solution
Now, whenever we compare any two solutions with some same or completely different parameters we use;
\[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\]
where, 1 represents the first condition and 2 represents the second condition.
Now, illustration;
Given that,
Density of acid solution = 1.8 g/ml = 1800 g/L
Thus, mass of ${{H}_{2}}S{{O}_{4}}$ present = $1800\times \dfrac{90}{100}=1620g$
Number of moles of ${{H}_{2}}S{{O}_{4}}$$\begin{align}
  & =\dfrac{actual-mass}{molar-mass} \\
 & =\dfrac{1620}{98}=16.53mol \\
\end{align}$
Thus, the molarity of the first solution is 16.53 M.
Now, to prepare 1 litre of 0.2 M solution,
\[\begin{align}
  & {{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}} \\
 & 16.53\times {{V}_{1}}=0.2\times 1 \\
 & {{V}_{1}}=\dfrac{0.2}{16.53}=0.01209L \\
\end{align}\]
Thus, the volume of acid to be used is 12.09 ml $\approx $ 12.1 ml.

Therefore, option (C) is correct.

Note:
Do note to use proper units while solving and forming new units in such problems. Also, do not get confused between molarity (M) and molality (m).
When, we compare two solutions with respect to their molarities; do note to use the formula as \[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\] .