Question

A solution containing 102 g of NaNO3 in 90.0 g H2O at 50 ∘C is cooled to 20 ∘C. Use the solubility data from the table below.

A) How many grams of NaNO3 remain in solution at 20 ∘C ?
B) How many grams of solid NaNO3 came out of solution after cooling?
if possible break it down step by step for me please. Thank you <3 :)

Answer 1

At 20 ℃, 79.2g of NaNO3 will remain dissolved in the solution, and 22.8g of NaNO3 will precipitate out of the solution.

When a solution of NaNO3 is cooled from 50 ℃ to 20 ℃, we must refer to the solubility data to determine how much solute remains in solution.

To answer part A, we need to look at the solubility of NaNO3 at 20 ℃.

Assuming the solubility at 20 ℃ is 88g per 100g of water, since we have 90.0g of water, the amount of NaNO3 that can remain in solution is:

(88g/100g) × 90.0g = 79.2g

To answer part B, we need to calculate how much NaNO3 will come out of solution after cooling based on the initial amount that was in solution (102g) and the amount that remains in solution (79.2g). The precipitated NaNO3 is:

102g - 79.2g = 22.8g

Hence, at 20 ℃, 79.2g of NaNO3 remains in solution, and 22.8g of NaNO3 will crystallize out of the solution.