Home
Class 11
CHEMISTRY
From 200 mg of CO(2), 10^(21) molecules...

From 200 mg of `CO_(2), 10^(21)` molecules are removed. How many moles of `CO_(2)` are left?

Text Solution

Verified by Experts

Total no. Of moles of `CO_(2) = ("wt.in g")/("mol. Wt")`
`=0.2/44 =0.00454`
No. Of moles removed `=(10^(21))/(6.022 xx 10^(23)) = 0.00166`
No. Of moles of `CO_(2)` left `=0.00454 - 0.00166=0.00288`
Promotional Banner

Similar Questions

Explore conceptually related problems

One mole of 'CO_2' contains

a) Calculate the mass of 112 L CO_2 gas kept at STP (molecular mass = 44). b) How many molecules of CO_2 are present in it?

How many molecules of water are present in one mole of water?

How many molecules are present in 1g N_2 ?

The enzyme carbonic anhydrase catalyses the hydration of CO_(2) . This reaction: CO_(2) + H_(2) to H_(2)CO_(3) , is involved in the transfer of CO_(2) from tissues to the lungs via the bloodstream. One enzyme molecule hydrates 10^6 molecules of CO_2 per second. How many kg of CO_2 are hydrated in one hour in one litre by 1 xx 10^(-6) M enzyme?

The molecules mass of oxygen is 32: How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?