Question

Calculate the molarity of each of the following solution:
(a) ${\rm{30}}\;{\rm{g}}$ of ${\rm{Co}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$ in ${\rm{4}}{\rm{.3}}\;{\rm{L}}$ of solution
(b) ${\rm{30}}\;{\rm{mL}}$of ${\rm{0}}{\rm{.5}}\;{\rm{M}}$ ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ diluted to ${\rm{500}}\;{\rm{mL}}$.

Answer 1

Hint:
We know that the molar mass is needed for calculating or identifying the number of moles of the given substance. The molarity is the combination of substances’ mole and the volume of the solution.

Complete step by step solution
(a).Given, the mass of cobaltous nitrate hexahydrate that is ${\rm{Co}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \\$ is ${\rm{30}}\;{\rm{g}}$ and the volume of the solution is ${\rm{4}}{\rm{.3}}\;{\rm{L}} \\$.
The molar mass of ${\rm{Co}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \\$ is ${\rm{291}}{\rm{.04}}\;{\rm{g/mol}}$,
The number of moles of ${\rm{Co}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}$ can be calculated by using the formula given below.
${\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\;{\rm{mass}}}}$
Substitute all the respective values in the above equation.
$
{\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = \dfrac{{{\rm{30}}\;{\rm{g}}}}{{{\rm{291}}{\rm{.04}}\;{\rm{g/mol}}}}\\
 = 0.103\;{\rm{mol}}
$
The moles of cobaltous nitrate hexahydrate is ${\rm{0}}{\rm{.103}}\;{\rm{mol}}$.
The molarity of the solution can be calculated by using the formula given below.
${\rm{Molarity}} = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{moles}}}}{{{\rm{Volume}}}}$
Substitute all the respective values in the above equation.
$
{\rm{Molarity}} = \dfrac{{{\rm{0}}{\rm{.103}}\;{\rm{mol}}}}{{{\rm{4}}{\rm{.3}}\;{\rm{L}}}}\\
 = 0.0239\;{\rm{M}}\\
 \approx 0.024\;{\rm{M}}
$

Hence, the molarity of the solution in question (a) is ${\rm{0}}{\rm{.024}}\;{\rm{M}}$.

(b) Given, the volume of sulphuric acid that is ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ is ${\rm{30}}\;{\rm{mL}}$ and molarity of ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ is ${\rm{0}}{\rm{.5}}\;{\rm{M}}$.
The volume of the diluted solution is ${\rm{500}}\;{\rm{mL}}$.
The molarity of the solution can be calculated by using the formula given below.
${{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}} = {{\rm{M}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}} $
Where, ${{\rm{M}}_{\rm{1}}}$ is the molarity of sulphuric acid, ${{\rm{V}}_{\rm{1}}}$ is the volume of sulphuric acid, ${{\rm{M}}_{\rm{2}}}$ is the molarity of diluted solution, and ${{\rm{V}}_{\rm{2}}}$ is the volume of diluted solution.
Substitute all the respective values in the above equation.
$
{\rm{30}}\;{\rm{mL}} \times {\rm{0}}{\rm{.5}}\;{\rm{M}} = {{\rm{M}}_{\rm{2}}} \times {\rm{500}}\;{\rm{mL}}\\
\Rightarrow {{\rm{M}}_{\rm{2}}} = \dfrac{{15}}{{500}}\;{\rm{M}}\\
 = 0.03\;{\rm{M}}
$

Hence, the molarity of diluted solution In question (b) is $0.03\;{\rm{M}}$.

Note:
The dilution equation is very necessary for the determination of unknown molarity of the given aqueous solutions.