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Standard XII

Chemistry

Rate Constant

For the decom...

Question

For the decomposition of $H_{2} O_{2} (a q .)$ , it was found that $V_{O_{2}} (t = 15 m i n .)$ was 100 mL (at 0 $^{o} C$ and 1 atm) while $V_{O_{2}}$ (maximum) was 200 mL (at 0 $^{o} C$ and 2 atm). If the same reaction had been followed by the titration method and if $V_{K M n O_{4}}^{c M} (t = 0)$ had been 40 mL, what would $V_{K M n O_{4}}^{c M} (t = 15 m i n)$ have been?

30 mL

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25 mL

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20 mL

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15 mL

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Solution

The correct option is A 30 mL
200 ml at 2 atm corresponds to 400 ml at 1 atm. This is the maximum value
In 15 minutes, the volume is 100 ml.
Thus one fourth of the reaction is complete in 15 minutes.
Hence, the volume of $K M n O_{4}$ will be $\frac{3}{4} \times 40 = 30 m L$

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For the decomposition of H2O2(aq.), it was found that VO2(t=15min.) was 100 mL (at 0oC and 1 atm) while VO2 (maximum) was 200 mL (at 0oC and 2 atm). If the same reaction had been followed by the titration method and if VcMKMnO4(t=0) had been 40 mL, what would VcMKMnO4(t=15min) have been?

The correct option is A 30 mL200 ml at 2 atm corresponds to 400 ml at 1 atm. This is the maximum valueIn 15 minutes, the volume is 100 ml.Thus one fourth of the reaction is complete in 15 minutes.Hence, the volume of KMnO4 will be 34×40=30mL

The correct option is A 30 mL
200 ml at 2 atm corresponds to 400 ml at 1 atm. This is the maximum value
In 15 minutes, the volume is 100 ml.
Thus one fourth of the reaction is complete in 15 minutes.
Hence, the volume of $K M n O_{4}$ will be $\frac{3}{4} \times 40 = 30 m L$