A solution is made by mixing 49.g of chloroform CHCl3 and 73.g of acetyl bromide CH3COBr. Calculate the mole fraction of chloroform in this solution. Round your answer to 2 significant digits.

The mole fraction of chloroform in this solution is 0.41

Explanation:

To calculate the moles, we use the equation:

a) moles of

b) moles of

To calculate the mole fraction, we use the formula:

The mole fraction of chloroform in this solution is 0.41

pure substance: iron,copper

impure substance: water,charcoal

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%