#### Question

Calculate the molarity of each of the following solutions: **(a)** 30 g of Co(NO_{3})_{2}. 6H_{2}O in 4.3 L of solution **(b)**30 mL of 0.5 M H_{2}SO_{4} diluted to 500 mL.

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#### Solution

Molarity is given by:

`"Molarity" = "Moles of solute"/"Volume of solution in litre"`

(a)Molar mass of Co (NO_{3})_{2}.6H_{2}O = 59 + 2 (14 + 3 × 16) + 6 × 18

= 291 g mol^{−1}

∴Moles of Co (NO_{3})_{2}.6H_{2}O = `30/291`mol

= 0.103 mol

Therefore, molarity = `"0.103 mol"/"4.3L"`

= 0.023 M

(b) Number of moles present in 1000 mL of 0.5 M H_{2}SO_{4} = 0.5 mol

∴Number of moles present in 30 mL of 0.5 M H_{2}SO_{4} = `(0.5xx30)/1000` mol

= 0.015 mol

Therefore, molarity = `0.015/"0.5L"` mol

= 0.03 M

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Solution for 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 concept: Expressing Concentration of Solutions. For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science), PUC Karnataka Science