4.2 g of a metallic carbonate MCO3 was heated in a hard glass tube and CO2 evolved was found to have 1120 mL of volume at STP. The Ew of the metal is
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a
12
b
24
c
18
d
15
answer is A.
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Detailed Solution
22400 mL=1 mol CO2=2 Eq of CO211200 mL=1/2 mol of CO2=1 Eq of CO2=1 Eq of CO32-Weight of metallic carbonate that would produce 1g equivalent or 22g or 11.2L of CO2 at STP would be its Ew.Ew of metallic carbonate = 4.2×1120011200=42gEw of metal = Ew of MCO3 - Ew of CO32-=42 - 30 = 12Ew of CO32-=602=30Alternatively:22400 mL = 1 mol of CO2 = 2 Eq CO32-11200 mL ⇒1 Eq of CO3-2∴1120 mL ⇒111200×1120=0.1 Eq of CO32-Eq of MCO3 = Eq of CO32-Weight Ew=WeightEw=0.1Eq∴0.1 = 4.2Ew of MCO3∴Eq of MCO3=42∴Ew of M=Ew of MCO3-Ew of CO32-= 42 - 30 = 12