Question

A 250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. The reaction is given. NaCl (aq)+AgNO, (aq)-AgCI(s)+NaNO, (aq) What is the concentration of the sodium chloride solution? (4) 7.1 Calculate the mass of the precipitate. 7.2 (4) 18

Answer 1

The mass of the precipitate that AgCl is 3.5803 g.

Step-by-step explanation:

a) To calculate the molarity of solution, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution =
`250 cm^3 =250 mL`

`1 cm^3= 1 ml`

Putting values in above equation, we get:

`\text{Molarity of solution}=(1.46g* 1000)/(58.5g/mol* 250)\\\\\text{Molarity of solution}=0.09982 M`

0.09982 M is the concentration of the sodium chloride solution.

b)
`NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)`

Moles of NaCl =
`(1.46 g)/(58.5 g/mol)=0.02495 mol`

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

`(1)/(1)* 0.02495 mol=0.02495 mol` of AgCl

Mass of 0.02495 moles of AgCl:

`0.02495 mol* 143.5 g/mol=3.5803 g`

The mass of the precipitate that AgCl is 3.5803 g.