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Question

21.2 g sample of impure Na2CO3 is dissolved and reacted with a solution of CaCl2, the weight of precipitate of CaCO3 is 10.0 g. Which of the following statements is/are correct ?

A
The % purity of Na2CO3 is 50%
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B
The percentage purity of Na2CO3 is 60%
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C
The number of moles of Na2CO3= that of CaCO3=0.1 mol
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D
The number of moles of NaCl formed is 0.1 mol
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Solution

The correct options are
A The % purity of Na2CO3 is 50%
C The number of moles of Na2CO3= that of CaCO3=0.1 mol
Molar mass of CaCO3=100 g mol1
Molar mass of Na2CO3=106 g mol1

Na2CO3+CaCl2CaCO3+2NaCl
(a) From stoichiometry,
moles of CaCO3=10100=0.1 mol
moles of Na2CO3= moles of CaCO3=12× moles of NaCl
mass of Na2CO3=0.1×106=10.6 g
% purity Na2CO3=10.621.2=100=50%

c) moles of CaCO3=10100=0.1 mol
(d) Moles of NaCl=2×0.1=0.2 mol

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