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100 g of an inorganic compound X.5H2O containing a volatile impurity was kept in an oven at 150 degree celcius for 60 minutes. the weight of residue after heating is 8g. the percentage of impurity in X was?

Respuesta :

              150°C
X*5H₂O    →    X + 5H₂O↑

m₀=100 g
m(X)=8 g
M(H₂O)=18.0 g/mol
n(H₂O)=5 mol

the mass of the water 
m(H₂O)=M(H₂O)n(H₂O)
m(H₂O)=18.0*5=90 g

the mass of the residue after removal of water
m'(X)=100-90=10 g

the percentage of impurity in X*5H₂O
w(imp)=[m'(X)-m(X)]/m₀

w(imp)=[10-8]/100=0.02 (2%)

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