Question

How many water molecules will be lost on dehydration of $0.684$ g of sucrose?

• A. $1.33\times {10}^{22}$
• B. $12\times {10}^{29}$
• C. $6.02\times {10}^{20}$
• D. $7.224\times {10}^{21}$

Answer 1

The correct option is A $1.33\times {10}^{22}$

The reaction for the dehydration of sucrose is $C_{12}{H}_{22}{O}_{11}\to 12C+11H_{2}O$.
Thus, from one molecule of sucrose, $11$ molecules of water are removed.
$0.684$ g of sucrose corresponds to $\frac{0.684}{342}=0.002$ moles which on dehydration will give $0.002\times 11=0.022$ moles of water.
This corresponds to $0.022\times 6.023\times {10}^{23}=1.33\times {10}^{22}$ water molecules.