Question

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Answer 1

Answer: (2) .

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Answer 2

Answer:

(2)  100 mL of M/10 AgNO3 + 100 mL of M/20 KI

Explanation:

It is clear that the precipitate formed is AgI

(AgNO3 + KI ------------->AgI + KNO3)

( 1 mole   1 mole             1 mole  1 mole)

The precipitate thus formed is of the size of a colloid particle.

AgI contains Ag(+) and I(-) ions.

If excess( I meant more than iodide ions) of SIlver ions(obtained from ionisable AgNO3) are present then the colloid becomes positively charged.

{molarity = number of moles x 1/volume in litres}

Let us check number of moles of AgNO3 and KI from each option,( Mind that you need to convert mL into L)

(1) Number of moles of AgNO3 is 0.1/10 = 0.01 moles.

   Number of moles of KI is 0.1/10= 0.01 moles.

Number of moles of AgNO3 = KI ( sol is neutral)

(2) Number of moles of AgNO3 is 0.1/10=0.01 moles.

    Number of moles of KI is 0.1/20 = 0.005 moles.

Number of moles of AgNO3 > KI ( sol is positive (ok))

(3) Number of moles of AgNO3 is 0.1/20 = 0.005 moles.

    Number of moles of KI is 0.1/10= 0.01 moles.

Number of moles of AgNO3 < KI ( sol is negative)

(4) Number of moles of AgNO3 is 0.1/20 = 0.005 moles.

    Number of moles of KI is 0.1/20 = 0.005 moles.

Number of moles of AgNO3 = Number of moles of KI (sol is neutral)

Hence the answer would be 100 mL of M/10 AgNO3 + 100 mL of M/20 KI