Question

Using the table, determine whether adding 11 g of NaNO3 to 25 g of H2O

will produce a saturated or unsaturated solution at 20 degrees Celsius:

Answer 1

Answer 2

Adding 11 g of
`NaNO_3` to 25 g of
`H_2O` will produce an unsaturated solution at 20 degrees Celsius.

According to the table, the solubility of
`NaNO_3` at 20 degrees Celsius is 88 g/100 g
`H_2O`. This means that 88 g of
`NaNO_3` can dissolve in 100 g of
`H_2O` to form a saturated solution at 20 degrees Celsius.

If we add 11 g of NaNO3 to 25 g of H2O, we are effectively adding less than the maximum amount that can be dissolved in 25 g of H2O at 20 degrees Celsius.

To find the maximum amount of NaNO3 that can be dissolved in 25 g of H2O, we can set up a proportion:

`$\frac{88 \text{ g NaNO3}}{100 \text{ g H2O}} = \frac{x \text{ g NaNO3}}{25 \text{ g H2O}} $$`

x = 88 x 25/100

= 22 g

Solving for x gives us 22 g of NaNO3.

Since 11 g is less than 22 g, adding 11 g of NaNO3 to 25 g of H2O will produce an unsaturated solution at 20 degrees Celsius.