JEE MAIN - Chemistry (2021 - 27th July Evening Shift - No. 16)
10.0 mL of 0.05 M KMnO4 solution was consumed in a titration with 10.0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is _________ $$\times$$ 10$$-$$2 g/L.
(Round off to the nearest integer)
(Round off to the nearest integer)
Answer
1575
Explanation
neq KMnO4 = neq H2C2O4 . 2H2O
or, $${{10 \times 0.05} \over {1000}} \times 5 = {{10 \times M} \over {1000}} \times 2$$
$$\therefore$$ Conc. of oxalic acid solution = 0.125 M
= 0.125 $$\times$$ 125 g/L = 15.75 g/L
= 1575 $$\times$$ 10$$-$$2 g/L
or, $${{10 \times 0.05} \over {1000}} \times 5 = {{10 \times M} \over {1000}} \times 2$$
$$\therefore$$ Conc. of oxalic acid solution = 0.125 M
= 0.125 $$\times$$ 125 g/L = 15.75 g/L
= 1575 $$\times$$ 10$$-$$2 g/L
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