CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

11.65 g of BaSO4 was obtained as a dry precipitate when 10 g of an impure sample of Na2SO4 was dissolved in water and treated with an excess of BaCl2. Calculate the percentage purity of the Na2SO4 sample. (Molar mass of BaSO4=233 g and Na2SO4=142 g)


A
38 %
loader
B
71 %
loader
C
22 %
loader
D
88 %
loader

Solution

The correct option is B 71 %
Na2SO4+BaCl2BaSO4+2NaCl
Moles of BaSO4=given massmolar mass=11.65233=0.05 moles
According to the reaction, 1 mole of BaSO4 precipitate is obtained from 1 mole of Na2SO4
0.05 moles of BaSO4 will be produced by 0.05 mole of Na2SO4
Amount of pure Na2SO4 required = 0.05 moles ×142 (gmole)=7.1 g
% purity =  mass of pure compound in sampletotal mass of impure sample×100=7.110×100
= 71 % 
 

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image