Question

The specific gravity of commercial sulphuric acid is $1.8$ and it is $98$ pure. The volume of this acid required for preparing $7.2\text{ }litres$ of decinormal sulphuric acid is:
A.$10ml$
B.$18ml$
C.$72ml$
D.$4ml$

Answer 1

Normality (N) is outlined because the range of gram equivalents of a matter gift per l of the answer. It’s given as:
$Normality=\dfrac{weight\text{ }of\text{ }solute\text{ }\left( in\text{ }grams \right)}{Equivalent\text{ }weight}$
If normality of one answer is given, that of alternative will be calculated mistreatment normality equation that is given as;
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
Specific gravity is the ratio of the density of the given substance to the density of water, Density of water is $1gm{{L}^{-1}}$

Complete step-by-step answer: To find normality of a $98$ solution of given sulphuric acid, ${{H}_{2}}S{{O}_{4}}$
Given molar mass of the ${{H}_{2}}S{{O}_{4}}=98grams$
Therefore ${{N}_{2}}$can be given as ${{N}_{2}}=\dfrac{1}{10}$
Thus, ${{M}_{2}}=\dfrac{{{N}_{2}}}{2}=\dfrac{\left( \dfrac{1}{10} \right)}{2}=\dfrac{1}{10}\times \dfrac{1}{2}=\dfrac{1}{20}$
${{V}_{2}}=7.2~$litre $98$ pure $98grams$​ ${{H}_{2}}S{{O}_{4}}$ in $100g$ solution,
Similarly for ${{M}_{1}}$ it is given by
${{M}_{1}}=\dfrac{98}{98}\times \dfrac{1000}{V}.................(here\text{ }1000\text{ }stands\text{ }for\text{ }1\text{ }litre\text{ }per\text{ }volume)...........(i)$
Now, $1ml-1.8gm$can be given by;
$100gm=\dfrac{1}{1.8}\times 100ml..............................\left( ii \right)$
Therefore, from $\left( i \right)\text{ }\And \text{ }\left( ii \right)$ we get ${{M}_{1}}$ as;
${{M}_{1}}=\dfrac{98}{98}\times \dfrac{1000}{100}\times 1.8$
${{M}_{1}}=\dfrac{1}{1}\times \dfrac{10}{1}\times 1.8$
${{M}_{1}}=1\times 10\times 1.8$
$\Rightarrow {{M}_{1}}=18ml$

Therefore, Option B is correct answer.

Note: Convert all the units for volume in either $l$ or $ml$ and use an equivalent unit throughout the calculation to avoid errors. Note and carefully follow steps to calculate volume of the solution from the particular gravity given.