Question

A 41-g sample of impure KClO3 (solubility = 7.1 g per 100 g H2O at 20°C) is contaminated with 13 percent of KCl (solubility = 25.5 g per 100 g of H2O at 20°C). Calculate the minimum quantity of 20°C water needed to dissolve all the KCl from the sample. (Assume that the solubilities are unaffected by the presence of the other compound.)

Answer 1

The minimum quantity of water at 20°C needed to dissolve all the KCl from a 41-g sample of impure KClO3, contaminated with 13 percent KCl, is approximately 20.9 g.

Step-by-step explanation:

The question involves calculating the minimum quantity of water needed to dissolve all the KCl from an impure KClO3 sample at 20°C. Given that the sample is contaminated with 13 percent KCl, we can determine how much KCl is present. Since the solubility of KCl is 25.5 g per 100 g of H2O, we can calculate the required amount of water to dissolve the KCl component fully.

First, we need to find the mass of the KCl in the impure sample:
Mass of KCl = 41 g × 13% = 5.33 g

Now, using the solubility of KCl, we can calculate the minimum amount of water needed:
Water needed for KCl = (5.33 g KCl) × (100 g H2O / 25.5 g KCl) <= 20.9 g of H2O (approximated for simplicity)

Therefore, at least 20.9 g of water at 20°C will be required to dissolve all the KCl present in the impure sample of KClO3.

Answer 2

Mass of water required to dissolve all KCl is 20.90 grams.

Step-by-step explanation:

Mass of the sample = 41 g

Percentage of KCl impurity = 13%

Mass of KCl =
`(13g)/(100)* 41 g=5.33 g`

Solubility of
`KCLO_3` = 7.1 grams per 100 grams of water at 20°C

Mass of
`KClO_3=41 g-5.33 g=35.67 g`

Solubility of
`KCL` = 25.5 grams per 100 grams of water at 20°C

Mass of
`KCl=5.33 g`

Then 1 gram of
`KCl` will dissolve in =
`(100)/(25.5)` grams of water

Mass of water required to dissolve 5.33 grams of
`KCl`:

`(100)/(25.5g)* 5.33=20.90 g`