Ncert Exercises & Solutions

2.3 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.

Molarity is given by:

$Molarity = \frac{Moles of solute}{Volume of solution in litre}$

(a) Molar mass of Co(NO₃)₂.6H₂O = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g mol⁻¹

$∴ Moles of Co (NO$ ₃)2.6H2O=30291mol=0.103 mol

Therefore, $molarity = \frac{0.103 mol}{4.3 L} = 0.023 M$

(b) Number of moles present in 1000 mL of 0.5 M H₂SO₄ = 0.5 mol

$∴$ Number of moles present in 30 mL of 0.5 M H₂SO_{4 $= \frac{0.5 \times 30}{1000} mol = 0.015 mol$}

Therefore, molarity $= \frac{0.015}{0.5 L} mol = 0.03 M$

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