Question

Calculate the molarity of each of the following solutions:
$\left( a \right)$ $30\,g$ of $Co{\left( {N{O_3}} \right)_2}.6{H_2}O$ in $4.3\,L$ of solution
$\left( b \right)$ $30\,mL$ of $0.5\,M$${H_2}S{O_4}$ dilute to $500\,mL$

Answer 1

Hint: First we have to know the molarity $\left( M \right)$, it is defined as the number of moles of solute dissolved in an amount of solvent (usually water). It is a concentration unit that describes how much of a substance is dissolved in solution.

Complete step by step answer:
We have to calculate the molarity $\left( M \right)$ for$\left( a \right)$,Let the no.of solute is $\left( n \right)$,weight of solute is $\left( w \right)$and molar mass is $\left( m \right)$.
Given,
Volume of solution $ = 4.3\,L$
Weight of solute $\left( {Co{{\left( {N{O_3}} \right)}_2}.6{H_2}O} \right)$ $ = 30\,g$
Molecular weight of $\left( {Co{{\left( {N{O_3}} \right)}_2}.6{H_2}O} \right)$ $ = 291\,g$
Molarity $M = \dfrac{{no.of\,moles\,of\,solute\left( n \right)}}{{Volume\,of\,solution\left( {in\,L} \right)}}$
And, no. of mole is defined as the weight of solute upon molar mass of solute $\left( n \right)$
$no.of\,mole\left( n \right) = \dfrac{{weight\,of\,solute\left( w \right)}}{{molar\,mass\left( m \right)}}$
First, calculation the no. of moles
$no.of\,mole\left( n \right) = \dfrac{{30\,g}}{{291\,g/mol}}$ …………………$(i)$
Now, for molarity $\left( M \right)$
Molarity $M = \dfrac{{\dfrac{{30}}{{291}}mol}}{{4.3\,L}}$………………… using $\left( i \right)$
Molarity $M = 0.024\,mol/L$
In $\left( b \right)$, We have to calculate the molarity $\left( M \right)$diluted solution
Given,
Volume of undiluted solution $\left( {{V_1}} \right)$ $ = 30\,mL$
Molarity of undiluted solution $\left( {{M_1}} \right)$$ = 0.5\,M$
Volume of diluted solution $\left( {{V_2}} \right)$$ = 500\,mL$
Molarity of diluted solution$\left( {{M_2}} \right)$$ = ?$
Calculate an unknown quantity where two solutions are proportional, calculating how much ingredient is containing in a different volume of the same concentration
$n{M_1}{V_1} = n{M_2}{V_2}$
$30 \times 0.5 = 500 \times {M_2}$
${M_2} = \dfrac{{30 \times 0.5}}{{500}}$
${M_2} = 0.03\,mol/L$
Hence, the molarity for $\left( a \right)$ and $\left( b \right)$ are $0.024\,mol/L$and $0.03\,mol/L$ respectively.

Note:
Molarity $\left( M \right)$indicates the number of moles per liter of solution $\left( {mol/Liter} \right)$ and is one of the most common units used to measure the concentration of a solution. Molarity can be used to calculate the volume or solvent or the amount of solute.
Molarity, molality and normality are all units of concentration in chemistry.
$1.$Molarity $\left( M \right)$ is defined as the no.of moles of solute per liter of solution.
$2.$Molality $\left( m \right)$ is defined as the no.of moles of solute per kilogram of solvent.
$3.$Normality $\left( N \right)$ is defined as the no.of equivalents per liter of solution.