The solubility of a compound in 35°C and 50°C is 60 and 80. 50g of saturated solution at 35°C is heated to 50°C.?

At 35°C :
Solubility = 60 g solute / 100 g of water = 60 g of the compound / 160 g of solution
Mass of solute in 80.50 g of saturated solution = (80.50 g) × (60/160) = 30.19 g
Mass of water in 80.50 g of saturated solution = (80.50 - 30.19) g = 50.31 g

At 50°C :
Solubility = 80 g solute / 100 g of water
Mass of solute dissolved in 50.31 g of water to form saturate solution = (50.31 g) × (80/100) = 40.25 g

Mass of extra solute required = (40.25 - 30.19) g = 10.06 g

can you post the question exactly as written here? It's impossible to answer this without knowing (1) units of 60 and 80, (2) solvent, (3) density of solution

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Supposing the units on "60" and "80" are grams of solute in 100 g of solvent.

So at 35°C, the mass ratio of saturated solute to solution is 60 g / (60 + 100) g = 0.375

When there is 50g of a solution at 35°C the mass of solute is:
(50 g) x 0.375 = 18.75 g solute
There is also 50 g - 18.75 g = 31.25 g of solvent.

At 50°C the ratio of saturated solute to solvent is: 80 g / (100) g = 0.80

When the solution containing 31.25 of solvent is heated to 50°C, the total amount of solute needed to make it saturated is:
(31.25 g) x (0.80) = 25 g solute

There was already 18.75 g of solute in it, so there needs to be:
25 g - 18.75 g = 6.25 g solute to be added.

(But that is supposing that the addition of 6.25 g of solid solute won't change the volume of the solution. You need to know the densities of the two saturated solutions in order to calculate a more exact answer.)