of potassium sulphate dissolves in  75gm . Water at  60∘C . Its Solubility at  60∘C .

To determine the solubility of potassium sulphate at 60∘C, we need to use the given information that a certain amount of potassium sulphate dissolves in 75gm of water at 60∘C.

Step 1: Convert the given amount of water from grams to milliliters. Since the density of water is 1g/mL, 75gm of water is equal to 75mL.

Step 2: Measure the amount of potassium sulphate that dissolves in the 75mL of water at 60∘C. Let's assume that x grams of potassium sulphate dissolves.

Step 3: The solubility of potassium sulphate at 60∘C is the amount of potassium sulphate that dissolves in 100mL of water at that temperature. To find this, we can use the concept of proportionality.

Since 75mL of water dissolves x grams of potassium sulphate, we can set up the following proportion:

75mL / x grams = 100mL / y grams

Where y represents the solubility of potassium sulphate at 60∘C.

Step 4: Solve the proportion for y. Cross-multiply and divide to find the value of y:

75mL * y = 100mL * x y = (100mL * x) / 75mL

Step 5: Simplify the expression:

y = (4/3) * x

Therefore, the solubility of potassium sulphate at 60∘C is (4/3) times the amount of potassium sulphate that dissolves in 75mL of water at that temperature.