Question

1.

A 0.4187 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 0.9536 g.

What is the mass percentage of chlorine in the original compound? ________%

2.

A student determines the calcium content of a solution by first precipitating it as calcium hydroxide, and then decomposing the hydroxide to calcium oxide by heating. How many grams of calcium oxide should the student obtain if her solution contains 49.0 mL of 0.593 M calcium nitrate?

_______g

3.

The bromide ion concentration in a solution may be determined by the precipitation of lead bromide.

• Pb²⁺(aq) + 2Br^-(aq) →→→→→PbBr₂(s)

A student finds that 20.36 mL of 0.7380 M lead nitrate is needed to precipitate all of the bromide ion in a 10.00-mL sample of an unknown. What is the molarity of the bromide ion in the student's unknown?

-------- M

Answer 1

Ans 1 :

Number of mol of AgCl = mass / molar mass

= 0.9536 g / 143.32 g/mol = 0.00665 mol

Each mol of AgCl has one mol of chloride ions , so here number number of mol of chlorine = 0.00665 mol

Mass of chlorine = mol x molar mass

= 0.00665 mol x 35.453 g/mol

= 0.236 g

Mass percentage of chlorine = ( mass of chlorine / mass of original compound) x 100

= ( 0.236 g / 0.4187 g) x 100

= 56.3 %