1.4 g of sample of chalk (CaCO3) containing clay as impurity were treated with excess of dilute HCl . volume of CO2 evolved when measured at 15°C and 768 mm pressure was 282 cm3 . calculate the % purity of the sample
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Dear frnd
Using ideal gas law, PV=nRT; Converting P in atm, V in litres and T in k n=0.01196~ 0.012
1 mole of CaCO3= 1 mole of CO2.
Therefore number of moles of CaCO3=0.012.
gram molecular mass of CaCO3= 0.012x100= 1.196.
% purity=100-( (1.4-1.196)/1.54x100)= 100-13.246=86.754
Using ideal gas law, PV=nRT; Converting P in atm, V in litres and T in k n=0.01196~ 0.012
1 mole of CaCO3= 1 mole of CO2.
Therefore number of moles of CaCO3=0.012.
gram molecular mass of CaCO3= 0.012x100= 1.196.
% purity=100-( (1.4-1.196)/1.54x100)= 100-13.246=86.754
niti117:
ans is 86.1
ur seems to b wrong
yaa i fault in calculation mistake ..
plz could u chk n let me know ur mistake....would b ur genriosity
plz chk i wrote the right answer .
no yaar its seriously wrong
u might have dad uaed the wrong way....perhaps
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