Question

Question #5e126

Answer 1

0.9139 liters

Explanation:

Mass of `C_6H_12O_6 = 72 + 12 + 64 = 148 gmol^-1`

Mole of glucose `rArr (1g)/(148gmol^-1) = 0.0068mol` of Glucose

Mole ratio between glucose and water is `1:6`

Then; water is `0.0068 xx 6 = 0.0408 mol` of water

Volume of water is `0.0408mol xx (22.4 L)/(1mol) = 0.9139 L`