Question

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.
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Answer 1

Given that mass of Co (NO3)2.6H2O  = 30 g

Volume of solution = 4.3 liter

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

Use the formula 



Moles of Co(NO3)2.6H2O = 30/291 = 0.103 mol

Use the formula 



Molarity of Co(NO3)2.6H2O = 0.103/4.3= 0.023 M

Answer 2

(a)

Given that mass of Co (NO3)2.6H2O = 30 g

Volume of solution = 4.3 liter

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

Use the formula

Moles of Co(NO3)2.6H2O = 30/291 = 0.103 mol

Use the formula

Molarity of Co(NO3)2.6H2O = 0.103/4.3= 0.023 M

(b)

Given that

Volume of H2SO4 (V1) = 30 ml)

Diluted volume of H2SO4 (V2) = 500 ml

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

Initial molarity M1= 0.5 M

We have to find M2

Use the formula

M2V2 = M1V1

plug the values we get

M2 × 500 = 0.5 × 30

M2 = 15/ 500 = 0.03 M