Question

$100$ g sample of clay (containing $19\%$ $H_{2}O$, $40\%$ silica and inert impurities as rest) is partially dried so as to contain $10\%$ $H_{2}O$. Which of the following is/are correct?

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

Answer 1

Answer: (A), (C)

The percentage of silica in it is $44.4\%$
The percentage of inert impurity in it is $45.6\%$

Amount of partially dried sample of clay $=100grams$

Then, amount of silica $=40grams$

Amount of water $=10grams$

(in partially dried sample)

Rest of the component i.e inert impurities $=[100-[40+10\left]\right]=50grams$

Total non water component $=90grams$ $[100-(40+50\left)\right]$

Original clay sample: Amount of water $=19grams$

Total non water component $=81grams$

Since, the proportion of non-water component remains same, Therefore,

Mass of Silica $=\frac{40}{81}\times 90$ $[=\frac{Amountofsilica}{Non{H}_{2}Ocomponentinoriginal}\times nonwatercomponentindriedsample]$

$=44.4$%

Mass of Inert impurity $=\frac{50}{90}\times 81$

$=45.6$%