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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

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 = `30/291`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 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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