How many molecules of `CO_(2)` will needed to obtain 1.8 g of glucose according to given reaction.
Reaction : `6CO_(2) + 6H_(2) O rarr C_(6) H_(12) O_(6) + 6O_(2)`
`[M. wt C_(6)H_(12)O_(6) = 180 g mol^(-1)]`
`(C=12, H=1, O=16)`

Options

A. `0.6xx6.022xx10^(23)`
B. `6xx6.022xx10^(23)`
C. `0.06xx6.022xx10^(23)`
D. `60xx6.022xx10^(23)`

Answer

A, B, C

Explanation

mole of glucose `=("mass ")/("molecular mass") = (1.8)/(180)`
` =0.01 ` mole glucose ...(1)
Reaction : `6CO_(2) + 6H_(2) O rarr C_(6)H_(12)O_(6) + 6O_(2)`
`:.` 6 mole `CO_(2)` is needed to get 1 mole glucose
`= (0.01xx6)/(1)`
`= 0.06` mole `CO_(2)` is needed ...(ii)
`=0.06xx6.022xx10^(23)` molecular of `CO_(2)`. ...(iii)