Bmat 2019 question 14

bmat 2019q14.pdf (89,4,KB)

Hey!
Is the answer B?

Hey Dorsa, I’m sorry for the delay and thanks for the feedack!
I also marked B, but the answer is A.

From the graph we can tell that the rate of reaction for Y is greater compared to X as it’s slope is steeper initially. However Y, produces 4/5 of the product that x does as observed by counting how many squares up on the Y axis is produced(measured as total volume of oxygen gas)
we can work out the moles of H2O2 using n=concentrationvolume=0.050.1=0.005
so, if curve Y produces 4/5 of the product=4/5*0.005=0.004
Now working from options A to E we can calculate the amount of moles from given concentration and volume and eliminate those options where no. of moles not equals 0.004 thus the answer is A
also, Manganese(IV)oxide is powdered as increased surface area would increase the rate of reaction.
Hope it helps.

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