5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure
water?

Explanation:

The formula for molality is:

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

Substitute the values into the formula.

Divide.

The molality is 2.8 moles per kilogram

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%