Suppose you have 0.7grams of unknown, containing 30% g KCl and 70% g KI. After d

Question

Suppose you have 0.7grams of unknown, containing 30% g KCl and 70% g KI. After dissolving in 100mL of water, you use 25.00 mL of it for the titration experiment. How many milliliter a of 0.070 M AgNO3 would you expect this mixture to consume in this titration. Suppose you have 0.7grams of unknown, containing 30% g KCl and 70% g KI. After dissolving in 100mL of water, you use 25.00 mL of it for the titration experiment. How many milliliter a of 0.070 M AgNO3 would you expect this mixture to consume in this titration.

Explanation / Answer

Answer – Given, mass of unknown = 0.70 g , mass percent of KCl = 30 % , mass percent of KI = 70 %

Now we need to calculate the mass of KCl and KI

We know,

Mass percent = mass / total mass *100 %

So, mass of KCl = 30 % *0.70 g / 100 %

                            = 0.21 g

Mass of KI = 70 % *0.70 g / 100 %

                  = 0.49 g

Now moles of KCl = 0.21 g / 74.55 g.mol-1

                               = 0.00282 moles

moles of KI = 0.49 g / g.mol-1

                   = 0.00295 moles

We know, KCl and KI both reacted with AgNO3 and form the precipitate as follow –

KCl + AgNO3 ---> AgCl + KNO3

KI + AgNO3 ---> AgI + KNO3

We know, total moles of 25 mL of KI + KCl = total moles of AgNO3

Moles of 100 mL of solution = 0.00577 moles (KCl+ KI)

We use 25 mL of solution = 0.00144

Moles of AgNO3 = 0.00144 moles

So, volume of AgNO3 = 0.00144 moles / 0.070 M

                                     = 0.0206 L

                                     = 20.6 mL

20.6 milliliter of 0.070 M AgNO3 would you expect this mixture to consume in this titration.

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