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Q.

5 g of K2SO4 was dissolved in 250 mL of solution. The volume of this solution that should be used so that 1.2 g of BaSO4 may be precipitated from BaCl2 is (molecular mass of K2SO4 =174 and BaSO4 = 233)

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a

55.2 mL

b

40.2 mL

c

44.8 mL

d

59.8 mL

answer is C.

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Detailed Solution

K2SO4+BaCl2⟶BaSO4+2KCl1mol 1 molMK2SO4=5×1000174×250=20174M×VLK2SO4= moles of BaSO420174×VL=1.2233VL=0.0448L=44.8mL

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5 g of K2SO4 was dissolved in 250 mL of solution. The volume of this solution that should be used so that 1.2 g of BaSO4 may be precipitated from BaCl2 is (molecular mass of K2SO4 =174 and BaSO4 = 233)