Question

2. 0.2500 g of a soluble chloride sample was dissolved in exactly 250.0 mL. It took exactly 40.00 mL of 0.01000 M Ag+ to titrate a 25.00 mL aliquot of the sample solution. Calculate the % Clin the sample.3. 100 mL of a river water sample took 9.30 mL of 0.01005 M Ag+ to titrate. Calculate the concentration of Clin ppm (µg/mL or mg/L) for the river water.

Answer 1

(2) The % of Cl in the sample is 56.8 %

(3) The concentration of Cl in ppm for the river water is 33.2 mg/L

Explanation :

Part 2 :

First we have to calculate the moles of
`Ag^+`

`\text{Moles of }Ag^+=Concentration* Volume=0.01M* 40.00mL=0.40mmol`

The chemical reaction will be:

`Ag^++Cl^-\rightarrow AgCl`

From the reaction we conclude that,

As, 1 mole of
`Ag^+` react with 1 mole of
`Cl^-`

So, 0.40 mmol of
`Ag^+` react with 0.40 mmol of
`Cl^-`

Now we have to calculate the moles of
`Cl^-` ion in 250 mL.

As, 25.00 mL contains moles of
`Cl^-` ion = 0.40 mmol

So, 250.0 mL contains moles of
`Cl^-` ion =
`(250.0)/(25.00)* 0.40=4mmol`

Now we have to calculate the mass of
`Cl^-` ion.

`\text{ Mass of }MgO=\text{ Moles of }MgO* \text{ Molar mass of }MgO`

Molar mass of
`Cl^-` ion = 35.5 g/mole

`\text{ Mass of }MgO=(4mmol)* (35.5g/mole)=142mg=0.142g`

Now we have to calculate the % of Cl in the sample.

`\% \text{ of Cl in sample}=(0.142g)/(0.25g)* 100=56.8\%`

Hence, the % of Cl in the sample is 56.8 %

Part 3 :

First we have to calculate the moles of
`Ag^+`

`\text{Moles of }Ag^+=Concentration* Volume=0.01005M* 9.30mL=0.0935mmol`

The chemical reaction will be:

`Ag^++Cl^-\rightarrow AgCl`

From the reaction we conclude that,

As, 1 mole of
`Ag^+` react with 1 mole of
`Cl^-`

So, 0.0935 mmol of
`Ag^+` react with 0.0935 mmol of
`Cl^-`

Now we have to calculate the mass of
`Cl^-` ion.

`\text{ Mass of }MgO=\text{ Moles of }MgO* \text{ Molar mass of }MgO`

Molar mass of
`Cl^-` ion = 35.5 g/mole

`\text{ Mass of }MgO=(0.0935mmol)* (35.5g/mole)=3.32mg`

Now we have to calculate the concentration of Cl in ppm.

`\text{Concentration of Cl in ppm}=(3.32mg)/(0.1L)=33.2mg/L`

Hence, the concentration of Cl in ppm for the river water is 33.2 mg/L