Question

The solubility of a salt is 20 g / 100 g water at 25 degree C. It increases 10% for every 5 degree C rise of temperature up to 50 degree C and then becomes constant. A solution of 26 g of the same salt in 100 g of water at 35 degree C would be​

Answer 1

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Answer 2

Given:

Solubility of salt - 20g/100g water ( 25 degree C)

To find:

Solubility of 26g of salt in 100g water at 35 degree C

Solution:

Solubility of salt is given by = $\frac{mass of solute}{mass of solvent}\times 100$

For the given salt at 25 degree C , solubility = $\frac{20}{100}\times 100$ $= 20$

A 5 degree C rise in temperature increases solubility by 10%,

At 30 degree C , solubility will be $= 20 + \frac{10}{100}\times 20$ = $22$

Further at 35 degree C , solubility will be  $= 22 + \frac{10}{100}\times 22$ = $24.2$

Now we have to find solubility of 26 g of same salt in 100g of water at 35 degree C.

From above calculation ; we know that , 20g/100g solution in water at 35 degree C has solubility $= 24.2$

Therefore , 26 g /100g in water, at 35 degree C will have solubility $=\frac{24.2\times 26}{20}$ = 31.46