Question

# 100 g sample of clay (containing 19% H2O, 40% silica and inert impurities as rest) is partially dried so as to contain 10% H2O. Which of the following is/are correct?

A
The percentage of silica in it is 44.4%
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B
The mass of partially dried clay is 90.0 g
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C
The percentage of inert impurity in it is 45.6%
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D
The mass of water evaporated is 10.0 g
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Solution

## The correct options are B The percentage of silica in it is 44.4% C The percentage of inert impurity in it is 45.6%Amount of partially dried sample of clay =100 gramsThen, amount of silica =40 gramsAmount of water =10 grams(in partially dried sample)Rest of the component i.e inert impurities =[100−[40+10]]=50 gramsTotal non water component =90 grams [100−(40+50)]Original clay sample: Amount of water =19 gramsTotal non water component =81 gramsSince, the proportion of non-water component remains same, Therefore, Mass of Silica =4081×90 [=Amount of silicaNon H2O component in original×non water component in dried sample]=44.4%Mass of Inert impurity =5090×81=45.6%

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