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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.

Respuesta :

Answer :

(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 [tex]Ag^+[/tex]

[tex]\text{Moles of }Ag^+=Concentration\times Volume=0.01M\times 40.00mL=0.40mmol[/tex]

The chemical reaction will be:

[tex]Ag^++Cl^-\rightarrow AgCl[/tex]

From the reaction we conclude that,

As, 1 mole of [tex]Ag^+[/tex] react with 1 mole of [tex]Cl^-[/tex]

So, 0.40 mmol of [tex]Ag^+[/tex] react with 0.40 mmol of [tex]Cl^-[/tex]

Now we have to calculate the moles of [tex]Cl^-[/tex] ion in 250 mL.

As, 25.00 mL contains moles of [tex]Cl^-[/tex] ion = 0.40 mmol

So, 250.0 mL contains moles of [tex]Cl^-[/tex] ion = [tex]\frac{250.0}{25.00}\times 0.40=4mmol[/tex]

Now we have to calculate the mass of [tex]Cl^-[/tex] ion.

[tex]\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO[/tex]

Molar mass of [tex]Cl^-[/tex] ion = 35.5 g/mole

[tex]\text{ Mass of }MgO=(4mmol)\times (35.5g/mole)=142mg=0.142g[/tex]

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

[tex]\% \text{ of Cl in sample}=\frac{0.142g}{0.25g}\times 100=56.8\%[/tex]

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

Part 3 :

First we have to calculate the moles of [tex]Ag^+[/tex]

[tex]\text{Moles of }Ag^+=Concentration\times Volume=0.01005M\times 9.30mL=0.0935mmol[/tex]

The chemical reaction will be:

[tex]Ag^++Cl^-\rightarrow AgCl[/tex]

From the reaction we conclude that,

As, 1 mole of [tex]Ag^+[/tex] react with 1 mole of [tex]Cl^-[/tex]

So, 0.0935 mmol of [tex]Ag^+[/tex] react with 0.0935 mmol of [tex]Cl^-[/tex]

Now we have to calculate the mass of [tex]Cl^-[/tex] ion.

[tex]\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO[/tex]

Molar mass of [tex]Cl^-[/tex] ion = 35.5 g/mole

[tex]\text{ Mass of }MgO=(0.0935mmol)\times (35.5g/mole)=3.32mg[/tex]

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

[tex]\text{Concentration of Cl in ppm}=\frac{3.32mg}{0.1L}=33.2mg/L[/tex]

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

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