Answer:
29 grams
Explanation:
At 20°C, a saturated solution of sodium nitrate (NaNO3) contains 86 grams of solute in 100 m of water.
This means the solubility of NaNO3 at 20°C is 86 g/100 ml of water
The solubility of NaNO3 at 50 °C is 115 grams of NaNO3/100 ml of water.
Therefore; to get the amount of sodium nitrate must be added to saturate the solution at 50°C;
We subtract the solubility at 20°C from the solubility at 50 °C
That is; 115 g/100 ml of water - 86 g/100 ml of water
= 29 grams of NaNO3
