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40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution. The liberated CO2 occupied 12.315 lit. at 1 atm & 300 K. The correct option is
  • a)
    Mass of impurity is 1 gm and metal is Be
  • b)
    Mass of impurity is 3 gm and metal is Li
  • c)
    Mass of impurity is 5 gm and metal is Be
  • d)
    Mass of impurity is 2 gm and metal is Mg
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
40 gm of a carbonate of an alkali metal or alkaline earth metal contai...
To calculate the volume of CO2 at STP condition,
Let say it will be V1
1×V / 273 = 1×12.315 / 300
​⇒V1=11.2L at STP
11.2L of CO2 at STP is equivalent to 0.5 mole of CO2. Suppose, Li is the metal
∴Li2CO3+2HCl⟶2LiCl+H2O+CO2
​∴  0.5 mole of Li2CO3 corresponds to 0.5×73.89 i.e. 37gm of Li2CO3
​∴  Wt. of impurity will be (40−37)gm i.e. 3gm
= 36.5×1 / 0.73=50
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Most Upvoted Answer
40 gm of a carbonate of an alkali metal or alkaline earth metal contai...
To calculate the molar amount of CO2 liberated, we use the ideal gas law:

PV = nRT

where P = 1 atm, V = 12.315 L, R = 0.082 L·atm/mol·K, and T = 273 K (standard temperature and pressure).

n = PV/RT = (1 atm)(12.315 L)/(0.082 L·atm/mol·K)(273 K) = 0.485 mol

Since one mole of the carbonate reacts with two moles of HCl to produce one mole of CO2, the molar amount of the carbonate is half of the molar amount of CO2:

n_carbonate = n_CO2/2 = 0.485/2 = 0.2425 mol

The molar mass of the carbonate can be calculated from its mass and molar amount:

molar mass = mass/molar amount = 40 g/0.2425 mol = 165.05 g/mol

Therefore, the carbonate is most likely calcium carbonate (molar mass = 100.09 g/mol) or magnesium carbonate (molar mass = 84.31 g/mol).
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Community Answer
40 gm of a carbonate of an alkali metal or alkaline earth metal contai...
B
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40 gm of a carbonate of an alkali metal or alkaline earth metal containing some inert impurities was made to react with excess HCl solution. The liberated CO2occupied 12.315 lit. at 1 atm & 300 K. The correct option isa)Mass of impurity is 1 gm and metal is Beb)Mass of impurity is 3 gm and metal is Lic)Mass of impurity is 5 gm and metal is Bed)Mass of impurity is 2 gm and metal is MgCorrect answer is option 'B'. Can you explain this answer?
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