Question

Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL.

Answer 1

$\textbf{molarity is the number of mole per }\\\textbf{unit volume in Litre}$

(a) weight of Co(NO3)2.6H2O = 30g
molar mass of Co(NO3)2.6H2O = 291 g/mol.
volume of solution = 4.3 L
$molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}$
so, number of moles of solute = given weight/molar mass
= 30/291 = 0.103 mol

molarity = 0.103mol/4.3L = 0.023M

(b) Given,
30 mL of 0.5 M H2SO4 diluted to 500 mL.
In 1000 mL of 0.5 M H2SO4, number of moles present is 0.5 mol.
∴ In 30 mL of 0.5 M H2SO4, number of moles present = 30 × 0.5/1000 = 0.015 mol
∴ molarity = number of moles present/volume
= 0.015mol/volume of solution
= 0.015/0.5L
= 0.03M

Answer 2

Answer :

(a)

Given :-

Mass of $Co(NO_{3})_{2}.6H_{2}O$ = 30 g

Volume of solution = 4.3 liter

Molar mass of $Co(NO_{3})_{2}.6H_{2}O$ = 59 + 2(14 + 3 × 16) + 6 (16 +1 × 2) = 291 g $mol^{-1}$

Molarity (M) = $\frac{W_{B}}{M_{B}}$ × $\frac{\text{1}}{\text{ Volume of solution in L}}$

⇒ M = $\frac{30}{294} \times \frac{1}{4.3} = \frac{10}{97} \times \frac{10}{43} = \frac{100}{4171}$

⇒ M = 0.024 mol $L^{-1}$

(b)

$M_{1} V_{1} = M_{2} V_{2}$

⇒ 0.5 × 30 ml = $M_{2}$ × 500

⇒ $M_{2} = \frac{15}{500} = \frac{3}{100}$

⇒ $M_{2}$ = 0.03 mol $L^{-1}$