Question

HCl solution (50 % by weight and specific gravity 1.25${g}/{ml}\;$) is used to prepare one litre N HCl. Calculate the volume of sample used for this purpose:
(A) 73 ml
(B) 58.4 ml
(C) 29.2 ml
(D) 36.5 ml

Answer 1

Hint: From the specific gravity of solution we can find the density of the solution. As we know, density is obtained by dividing mass of the substance by volume. Thus by dividing the mass with density, we can calculate the volume of the sample.

Complete step by step solution:
- Let's begin with the idea of specific gravity. We can explain it as a ratio between the density of an object and a reference substance at the same temperature. In most cases the reference substance is taken as water.
- We are given that the specific gravity of the given solution is $1.25$${g}/{ml}\;$. Hence, we can write the relation between density (ρ) and specific gravity as follows
\[1.25=\dfrac{{{\rho }_{HCl}}}{{{\rho }_{{{H}_{2}}O}}}\]
\[{{\rho }_{HCl}}=1.25\times {{\rho }_{{{H}_{2}}O}}\]
The specific gravity of water is usually taken as 1 ${g}/{ml}\;$.Lets substitute this in the above equation,
\[{{\rho }_{HCl}}=1.25\times 1\]
\[{{\rho }_{HCl}}=1.25{g}/{ml}\;\]
As we know, density is obtained by dividing mass of the substance by volume and we can write as follows
\[V=\dfrac{m}{\rho }\]
 The mass of HCl is 36.5 and we got the value of density as 1.25${g}/{ml}\;$.Lets substitute this values in the above equation
\[V=\dfrac{36.5}{1.25}=29.2ml\]
Therefore the volume of sample solution of HCl used for this purpose is 29.2 mL.

Thus the answer is option (C) 29.2 ml

Note: Keep in mind that the normality is the molecular weight divided by the grams per equivalent in a given volume. For a 1 N solution we need 1 equivalent/liter. For HCl the equivalent weight is 36.5 grams. Therefore, for making a one normal solution, 36.5${g}/{litre}\;$of HC1 is needed. This is the reason why we take the mass of HCl as 36.5 grams.