25mL of a solution of Fe2- ions was titrated with a solution of the oxidising agent Cr2O2-7- 32-45mL of 0-0153M K2Cr2O7 solution was required- What is the molarity of the Fe2- solution-

The correct option is A 0.1192 M Fe^+2 + Cr_2 O_7^2 → Fe^+3 + Cr^+3 Balance above equation 6Fe^+2 + Cr_2O_7^2 + 14 H^+ → 6Fe^+3 + 2 Cr^+3 + 7 H_ ...

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Standard XII

Chemistry

Molarity

25mL of a sol...

Question

$25$ mL of a solution of $F e^{2 +}$ ions was titrated with a solution of the oxidising agent $C r_{2} O_{7}^{2 -}$ , $32.45$ mL of $0.0153$ M $K_{2} C r_{2} O_{7}$ solution was required. What is the molarity of the $F e^{2 +}$ solution?

0.4322 M

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0.1192 M

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0.6752 M

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0.3439 M

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Solution

The correct option is A $0.1192 M$
$F e^{+ 2} + C r_{2} O_{7}^{2 -} \to F e^{+ 3} + C r^{+ 3}$
Balance above equation
$6 F e^{+ 2} + C r_{2} O_{7}^{2 -} + 14 H^{+} \to 6 F e^{+ 3} + 2 C r^{+ 3} + 7 H_{2} O$
Milli mole of $C r_{2} O_{7}^{2 -} = 32.45 \times 0.0153 = 0.496$
$1$ millimole of $C r_{2} O_{7}^{2 -}$ oxidise $6$ millimole of $F e^{+ 2}$
$0.496$ $0.496 \times 6$
$0.496 \times 6 = M \times 25 m l$
$M = \frac{0.496 \times 6}{25} = 0.1192 M$

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Similar questions

25

mL of a solution of

F e^{2 +}

ions was titrated with a solution of the oxidising agent

C r_{2} O_{7}^{2 -} . 32.45

mL of

0.0153

K_{2} C r_{2} O_{7}

solutin was required. What is the molarity of the

F e^{2 +}

solution?

Q. When

10 m L

of an aqueous solution of

F e^{2 +}

ions was titrated in the presence of dil

H_{2} S O_{4}

using diphenylamine indicator,

15 m L of

0.02 M

solution of

K_{2} C r_{2} O_{7}

was required to get the end point. The molarity of the solution containing

F e^{2 +}

ions is

x \times 10^{- 2} M

. The value of

x

is ___. (Nearest integer)

Q. A mixture of

F e O

and

F e_{2} O_{3}

is completely reacted with

100 mL

0.25 M

acidified

K M n O 4

solution. The resultant solution was then titrated with

Z n

dust which converted the

F e^{3 +}

ions to

F e^{2 +}

ions. The

F e^{2 +}

ions required

1000 mL

0.10 M

K_{2} C r_{2} O_{7}

solution. Find out the weight percentage of

F e_{2} O_{3}

in the mixture.

Q. A

10 g

sample of

C u S

and

C u_{2} S

was treated with

100 m L

1.25 M K_{2} C r_{2} O_{7}

to produce

C r^{3 +}, C u^{2 +}

and

S O_{2}

. The excess oxidant was reacted with

50 m L

F e^{2 +}

solution.

25 m L

of the same

F e^{2 +}

solution required

0.875 M K M n O_{4}

under acidic condition, the volume of

K M n O_{4}

used was

20 m L

. Find the percentage of

C u_{2} S

in the sample.
Given: molar mass of

C u

63.5 g m o l^{- 1}

Q. The following titration method is given to determine the total content of the species with variable oxidation states. Answer the question given at the end of it.

A quantity of

25.0

mL of solution containing both

F e^{2 +}

and

F e^{3 +}

ions is titrated with

25.0

mL of

0.02 M

K M n O_{4}

(in dilute

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

ions are oxidised to

F e^{3 +}

ions. Next

25

mL of the original solution is treated with

Z n

metal. Finally, the solution requires

40.0

mL of the same

K M n O_{4}

solution for oxidation to

F e^{3 +}

{M n O_{4}}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

The molar concentration of

F e^{2 +}

in the original solution is :

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25mL of a solution of Fe2+ ions was titrated with a solution of the oxidising agent Cr2O2−7, 32.45mL of 0.0153M K2Cr2O7 solution was required. What is the molarity of the Fe2+ solution?

The correct option is A 0.1192 MFe+2+Cr2O2−7→Fe+3+Cr+3Balance above equation6Fe+2+Cr2O2−7+14H+→6Fe+3+2Cr+3+7H2OMilli mole of Cr2O2−7=32.45×0.0153=0.4961 millimole of Cr2O2−7 oxidise 6 millimole of Fe+20.496 0.496×60.496×6=M×25mlM=0.496×625=0.1192M

$25$ mL of a solution of $F e^{2 +}$ ions was titrated with a solution of the oxidising agent $C r_{2} O_{7}^{2 -}$ , $32.45$ mL of $0.0153$ M $K_{2} C r_{2} O_{7}$ solution was required. What is the molarity of the $F e^{2 +}$ solution?