Question

A 10.151 g sample of window cleaner containing ammonia was diluted with 39.466 mL of water. Then 4.373 g of solution was titrated with 14.22 mL of 0.1063 M HCl to reach a bromocresol green end point. Find the weight percent of NH3 (FM 17.031) in the cleaner.

Answer 1

Answer:

2.878 (w/w)%

Explanation:

The reaction of NH3 with HCl is:

NH3 + HCl → NH4Cl

That means 1 mole of ammonia reacts per mole of HCl. The moles of HCl that reacts = Moles of ammonia are:

14.22mL = 0.01422L * (0.1063mol HCl / L) = 0.001512 moles HCl = Moles ammonia

In mass:

0.001512 moles NH3 * (17.031g / mol) = 0.02574g of NH3 in the solution.

The concentration in weight percent of the diluted solution is:

0.02574g of NH3 / 4.373g * 100 = 0.5887 (w/w)%

This is the concentration of the diluted solution that was diluted from 10.151g to 10.151g + 39.466g = 49.617g

-Assuming density of water = 1g/mL-

The solution was diluted:

49.617g / 10.151g = 4.888 times.

That means the concentration of NH3 in the original cleaner is:

0.5887 (w/w)% * 4.888 =

2.878 (w/w)%