Question

When 180 g of glucose is subjected to combustion, the volume of CO2 liberated at STP is:
(A) 22.41
(B) 67.21
(C) 441
(D) 134.41​

Answer 1

Answer:

D)134.41

Explanation:

The formula of glucose is C6H12O6.

molecular mass=12×6+1×12+16×6=180g

Now, when glucose breakdown,6 molecule of CO2 produce means that from 1 molecules of glucose 6 molecule of CO2 produce.

We know that,

At STP, one mole occupy 22.4L

So, the volume of CO2 => 6×22.4L => 134.41 L