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# Solution - 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 - Expressing Concentration of Solutions

ConceptExpressing Concentration of Solutions

#### 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.

#### Solution

Molarity is given by:

"Molarity" = "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 (NO3)2.6H2O = 30/291mol

= 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 H2SO4 = 0.5 mol

∴Number of moles present in 30 mL of 0.5 M H2SO4 = (0.5xx30)/1000  mol

= 0.015 mol

Therefore, molarity = 0.015/"0.5L" mol

= 0.03 M

Is there an error in this question or solution?

#### APPEARS IN

NCERT Chemistry Textbook for Class 12 Part 1
Chapter 2: Solutions
Q: 3 | Page no. 37

#### Reference Material

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
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