Question

a 50 g sample of impure kclo3 solubility - 7.1 g per 100 g h2o at 20 degree c is contaminated with 10 percent of kcl solubility - 25.5 g per 100 g of h2o at 20 degree c. calculate the minimum quantity of 20 degree c water needed to dissolve all the kcl from the sample. how much kclo3 will be left after this treatment? assume that the solubilities are unaffected by the presence of the other compound.

Answer 1

First, let's calculate the amount of KCl in the 50 g sample of impure KClO3:

The sample is 10% KCl, so there is 0.1 x 50 g = 5 g of KCl in the sample.

To dissolve all of the KCl from the sample, we need to find the amount of water required that can dissolve 5 g of KCl at 20°C:

The solubility of KCl at 20°C is 25.5 g per 100 g of water, so to dissolve 5 g of KCl, we need (5 g / 25.5 g/100g) x 100 g = 19.61 g of water.

So, we need at least 19.61 g of water to dissolve all the KCl from the sample.

Next, let's calculate how much KClO3 will be left after we remove all the KCl from the sample:

The original sample was 50 g, and 10% of it is KCl, so the amount of KClO3 in the sample is 90% of 50 g = 45 g.

If we remove 5 g of KCl from the sample, the remaining weight of the sample will be 50 g - 5 g = 45 g.

Therefore, the remaining weight of KClO3 in the sample will also be 45 g.

So, after removing all the KCl from the sample, we will have 45 g of KClO3 left.