1-2 gm of a commercial sample of oxalic acid H 2 C 2 O 4- 2 H 2 O was dissolved in 200 mL of water- If 10 mL of this sample required 8-5 mL of N -10 KMnO 4- then percentage purity of sample of oxalic acid is-A- 67-2 -B- 54-7 -C- 89-25 -D- 96-5 -

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Byju's Answer

Standard XII

Chemistry

Conductance of Electrolytic Solution

1.2 gm of a c...

Question

1.2 gm of a commercial sample of oxalic acid $H_{2} C_{2} O_{4} . 2 H_{2} O$ was dissolved in 200 mL of water. If 10 mL of this sample required 8.5 mL of $\frac{N}{10} K M n O_{4},$ then percentage purity of sample of oxalic acid is:

54.7%

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67.2%

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89.25%

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96.5%

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Solution

The correct option is C 89.25%
At equivalence point
$E q ._{o x a l i c a c i d} = (E q .)_{K M n o_{4}} (N \times V)_{o x a l i c a c i d} = (N \times V)_{K M n o_{4}}$
Let the amount of pure oxalic acid sample is x gm
$\frac{x \times 2 \times 1000}{126 \times 200} \times 10 = 8.5 \times \frac{1}{10} x = 1.07 g m % p u r i t y o f s a m p l e = \frac{1.07}{1.2} \times 100$
= 89.25%

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1.2 gm of a commercial sample of oxalic acid H2C2O4.2H2O was dissolved in 200 mL of water. If 10 mL of this sample required 8.5 mL of N10KMnO4, then percentage purity of sample of oxalic acid is:

The correct option is C 89.25%At equivalence point Eq.oxalicacid=(Eq.)KMno4(N×V)oxalic acid=(N×V)KMno4 Let the amount of pure oxalic acid sample is x gm x×2×1000126×200×10=8.5×110x=1.07 gm%purity of sample =1.071.2×100 = 89.25%

1.2 gm of a commercial sample of oxalic acid $H_{2} C_{2} O_{4} . 2 H_{2} O$ was dissolved in 200 mL of water. If 10 mL of this sample required 8.5 mL of $\frac{N}{10} K M n O_{4},$ then percentage purity of sample of oxalic acid is: