Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is:
(The formula for Mohr's salt is $F e S O_{4} . (N H_{4})_{2} S O_{4} .6 H_{2} O$ )

Question

Consider a titration of potassium dichromate solution with acidified Mohr-s salt solution using diphenylamine as indicator- The number of moles of Mohr-s salt required per mole of dichromate is-3456

Toppr Admin · Accepted Answer

Cr2O2-x2212-7- Fe 2-x2192- Fe 3-Cr 3-n factor of Cr2O2-x2212-7-6n factor of Fe2-1So- to reduce one mole of dichromate- 6 moles of Fe2- are required-xA0-Hence- the correct option is D

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is:
$3$
$4$
$5$
$6$

$C r_{2} O_{7}^{2 -} +$ $F e$ $^{2 +} \to$ $F e$ $^{3 +} + C r$ $^{3 +}$

$n$ factor of $C r_{2} O_{7}^{2 -} = 6$

$n$ factor of $F e^{2 +} = 1$

So, to reduce one mole of dichromate, $6$ moles of $F e^{2 +}$ are required.

Hence, the correct option is $D$

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is:3456

Cr2O2−7+ Fe 2+→ Fe 3++Cr 3+n factor of Cr2O2−7=6n factor of Fe2+=1So, to reduce one mole of dichromate, 6 moles of Fe2+ are required. Hence, the correct option is D

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is:
$3$
$4$
$5$
$6$

$C r_{2} O_{7}^{2 -} +$ $F e$ $^{2 +} \to$ $F e$ $^{3 +} + C r$ $^{3 +}$

$n$ factor of $C r_{2} O_{7}^{2 -} = 6$

$n$ factor of $F e^{2 +} = 1$

So, to reduce one mole of dichromate, $6$ moles of $F e^{2 +}$ are required.

Hence, the correct option is $D$