Question

A solution of $HCl$ acid contains $36\%$ by mass. Calculate the molal and molar concentration of $HCl$ solution? The density of $HCl$ solution is $1.18\;gm{L^{ - 1}}$ . Consider there is $190\;mL$ of solution.

Answer 1

To solve this question we have to know about molarity and molality of the substance. These are the two ways of expressing the concentration of the solution. Thus after knowing their formulas we will substitute the values and find the required answer.
$molarity = \dfrac{{moles\;of\;solute\left( {mol} \right)}}{{volume\;of\;solution\left( L \right)}}$
And, $molality = \dfrac{{moles\;of\;solute\left( {mol} \right)}}{{mass\;of\;solvent\left( {kg} \right)}}$ .

Complete answer:
Molarity which is alternatively known as Molar concentration is the measure of concentration of the chemical substance, which is the amount of a compound in a certain volume of solution. It is also defined as the amount of solute in moles present per litres of solution.
$molarity = \dfrac{{moles\;of\;solute\left( {mol} \right)}}{{volume\;of\;solution\left( L \right)}}$
Thus to get an answer, we will first take $1mL$ volume whose mass is $1.18g$ and then find the moles of solute which is:
$molarity = \dfrac{{\dfrac{{1.18gm{L^{ - 1}} \times 1mL}}{{36.46gmo{l^{ - 1}}}} \times 36\% }}{{1mL \times {{10}^{ - 3}}Lm{L^{ - 1}}}}$
$\Rightarrow 11.65\;mol{L^{ - 1}}$
But this is slightly more concentrated than the conc. Hydrochloric we use in the lab which is:
$\left[ {HCl} \right] = 10.6\;mol{L^{ - 1}}$
Molality which is alternatively known as Molal concentration is the measure of concentration of the chemical substance, which is the amount of the compound in a certain mass of solvent. It can also be defined as the amount of solute in moles present per kilogram of the solvent.
$molality = \dfrac{{moles\;of\;solute\left( {mol} \right)}}{{mass\;of\;solvent\left( {kg} \right)}}$
Again for the answer we will take $1mL$ volume whose mass is $1.18g$ which is $0.425g$ with respect to the acid and $0.7552g$ with respect to water solvent. So,
$molality = \dfrac{{\dfrac{{0.425g}}{{36.46gmo{l^{ - 1}}}}}}{{0.7552g \times {{10}^{ - 1}}kg\;{g^{ - 1}}}}$
$\Rightarrow 15.4\;mol\;k{g^{ - 1}}$

Note:
Both molarity and molality are used to express the concentration of the solution. But the molarity of substance depends upon temperature whereas the molality of the substance does not depend upon temperature.