In: Chemistry
A solution containing 95.9 g of NaNO3 in 90.0 gH2O at 50 ∘C is cooled to 20 ∘C. Use the solubility data from the table below.
Solubility (g/100gH2O) | ||
Substance | 20∘C | 50∘C |
KCl | 34.0 | 42.6 |
NaNO3 | 88.0 | 114.0 |
C12H22O11 (sugar) | 203.9 | 260.4 |
1. How many grams of NaNO3 remain in solution at 20 ∘C?
2. How many grams of solid NaNO3 came out of solution after cooling?
From the solubility table ,
100g of water can dissolve 114.0g of NaNO3 at 50C
Thus 90.0g of water can dissolve = 90.0g x 114.0g / 100g
= 102.6g
Thus when we dissolve 95.9g of NaNO3 completely dissolves in 90.0g of NaNO3 at 50C
When it is cooled to 20C , the solubility decreases , thus some solid coes out.
At 20C
100g of H2O dissolves 88.0g of NaNO3
90.0g of water can dissolve = 90x88 /100
=79.2 g
Thus 79.2g of NaNO3 remains soluble and the rest (95.9g - 79.2 g) will be coming out as precipitate.
Q1) the amount of NaNO3 that remains in solution at 20C = 79.2 g
Q2) the amont of solid NaNO3 that came out of solution = (95.9g - 79.2 g)
= 16.7g