Question

A 30% solution of X at 298 K is marked as a saturated solution of X. What is the solubility of X at 298 K?​

Answer 1

Answer:

The conductivity (K) of a saturated solution of AgBr at 298 K is 8.5x10-7 Scm . ... and ce are 62 and 78 Scmmoll, respectively the Ksp of AgBr.

Answer 2

The solubility of X at 298 K is 42.85%

GIVEN

A 30% solution of X at 298 K is marked as a saturated solution of X.

TO FIND

The solubility of X at 298 K

SOLUTION

We can simply solve the above problem as follows;

We know that when the solute concentration is equal to the solubility of solution, the solution is marked as a saturated solution.

Mass% of solute = 30%

This means,

30 grams of solute is present in 100 grams of solution.

Mass of solvent = Mass of solution - Mass of solute.

= 100-30

= 70 grams

$solubility\% = \frac{mass \: of \: solute \: }{mass \: of \: solvent} \times 100$

Putting the values in the above formula we get,

$= \frac{30}{70} \times 100$

= 42.85%

Hence, The solubility of X at 298 K is 42.85%

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