Question

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

Answer 1

Hint : Here you should recall the basic concept of molarity. A solution's molarity tells you about the number of moles of solute that is present in 1 L of solution. You can say:
$Molarity(M) = \dfrac{{moles{\text{ }}of{\text{ solute}}}}{{1{\text{ }}litre{\text{ of }}solution{\text{ }}}}$ .

Complete Step By Step Answer:
We know that in order to calculate a solution's molarity, you should know the number of moles of solute that is present in 1 L i.e. 1000 mL of solution.
We can use the molar mass of a compound to determine the number of moles present in the sample so number of moles of a compound can be calculated by applying the following formula:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}mass(gmo{l^{ - 1}})}}$
Now, let us calculate the molarity of each of the given solutions one by one:
(a) Mass of $Co{\left( {N{O_3}} \right)_2}.6{H_2}O$ = $30{\text{ }}g$ (Given)
Volume of solution = $4.3{\text{ }}L$ (Given)
We know that the molar mass of any compound can be found out by adding the relative atomic masses of each element present in that particular compound. Here molar mass of the compound can be calculated as stated below:
$Molar{\text{ }}mass{\text{ }}of{\text{ }}Co{\left( {N{O_3}} \right)_2}.6{H_2}O = (1 \times Co) + (2 \times N) + (12 \times O) + (12 \times H)$
$= (1 \times 58.9) + (2 \times 14) + (12 \times 16) + (12 \times 1) = 290.9u$
Substituting the values in the number of moles formula, we get:
$Number{\text{ }}of{\text{ }}moles = \dfrac{{30g}}{{290.9gmo{l^{ - 1}}}} = 0.103mol$
Now substituting the values in molarity formula, we get:
$Molarity = \dfrac{{0.103}}{{{\text{4}}{\text{.3 }}}} = 0.023M$
(b) In the question we are provided with the following information:
$30{\text{ }}mL$ of $0.5M$ ${H_2}S{O_4}$ diluted to $500{\text{ }}mL$
In $1000{\text{ }}mL$ of $0.5M$ ${H_2}S{O_4}$ , number of moles = $0.5{\text{ }}mol$
So, in $30{\text{ }}mL$ of $0.5M$ ${H_2}S{O_4}$ , number of moles present = $\dfrac{{30 \times 0.5}}{{1000}} = 0.015{\text{ }}mol$
Now substituting the values in molarity formula, we get:
$Molarity = \dfrac{{0.015}}{{{\text{0}}{\text{.5 }}}} = 0.03M$

Note :
Don’t get confused between molarity and molality. Molarity is the number of moles of solute per volume of solution (litres) while Molality is the number of moles of solute per weight of solvent (kilogram). These two have a relationship as below:
Molality = (density of the solution − molarity) × molecular weight of solute molarity.