KVPY 2019 SB SX Exam Previous Papers

Question : 109 of 120

Marks: +1, -0

n a titration experiment, 10 mL of an

FeCl_2

solution consumed 25 mL of a standard

K_2Cr_2O_7

solution to reach the equivalent point. The standard

K_2Cr_2O_7

solution is prepared by dissolving 1.225 g of

K_2Cr_2O_7

in 250 mL water. The concentration of the

FeCl_2

solution is closest to
[Given : molecular weight of

K_2Cr_2O_7

= 294 g mol

^{–1}

]

Solution:

Molarity of

K_2Cr_2O_4 = {1.225× 1000}/{294× 250}

= 0.0167

{Fe^{2+}}↙{VF= 1}+{Cr_2O_7^{-2}}↖{+6}↙{VF= 6}\ –––▸ Fe^{3+}Cr^{3+}↖{+3}

Mili Eq. Fe

^{2+}

= mili eq. of

Cr_2O_7^{2–}

[M × 10]1 = [0.016 × 25]6
M = 0.25
For FeCl

{_2}

concentration = 0.25 N

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