Question

In: Chemistry

A solution containing 95.9 g of NaNO3 in 90.0 gH2O at 50 ∘C is cooled to...

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?

Solutions

Expert Solution

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


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