SELECT * FROM question_mgmt as q WHERE id=1536 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

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.

Answer:

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 / 4.3 L

= 0.024 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   = (0.5 X 30 ) / 1000 mol

= 0.015 mol

Therefore, molarity = 0.015 mol / 0.5 L

= 0.03 M


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