Question

In an experiment 100 grams of dry soil is taken and water from a measuring cylinder is given to it. It is observed that the initial volume of water is 500 mL and the when the soil is completely wet and just percolate the volume of water remain in the cylinder is 450 mL. what is the percentage of water absorbed?

Answer 1

Given :-

Weight of the dry soil taken for the experiment=100 grams

We know that :-

The intital weight of water =U

The final weight of water =V

Formula for finding the volume of water absorbed by the soil:-

$= \tt{(U - V) \: \: ml}$

Formula for finding the weight of water absorbed by the soil:-

$= \tt{(U - V) \: \: grams}$

Which means :-

The volume of water absorbed by this dry soil:-

$=( 500 - 450) \: \: ml$

$=\bold{ 50 \: \: ml}$

Thus, the volume of soil absorbed by this dry soil=50 ml

The weight of water absorbed by this dry soil:-

$= (500 - 450) \: \: grams$

$=\bold{ 50 \: grams}$

Thus, the weight of water absorbed by this dry soil=50 grams

Formula for finding the percentage of water absorbed by soil:-

$= ( \frac{\tt{U - V}}{\tt{weight \: of \: soil}} ) \: \times \: \tt{ 100}$

$= \frac{500 - 450}{100} \times 100$

$= \frac{50}{100} \times 100$

$= \frac{5000}{100}$

$= \frac{5000 \div 100}{100 \div 100}$

$= \bold{50 \: \%}$

Thus, percentage of water absorbed =50%

Therefore, the percentage of water absorbed by the dry soil=50%