Question

What is the volume of water consumed during acid hydrolysis of 1.368 kg of sucrose?

[Given : molar masses of sucrose = 342, water = 18, density of water = 18, density of water = 1g/cm${}^3$]

• A. $0.072d{m}^3$
• B. $0.720d{m}^3$
• C. $0.18d{m}^3$
• D. $0.018d{m}^3$

Answer 1

The correct option is A $0.072d{m}^3$

$C_{12}{H}_{22}{O}_{11}+H_{2}O\stackrel{H^{+}}{\to }{C}_6{H}_{12}{O}_6+C_6{H}_{12}{O}_6$

1 mol of sucrose requires 1 mol of $H_{2}O$.

1.368 g of $C_{12}{H}_{22}{O}_{11}$ has $\frac{1.368\times {10}^3}{342}=4$ moles.

4 moles of sucrose requires the same number of moles of $H_{2}O$.

Molar mass of $H_{2}O=18$ g/mol

So, mass of $H_{2}O$ required $=18\times 4=72$ gm of $H_{2}O$.

The density of water $=1g/c{m}^3$.

So, volume of water required $=\frac{mass}{density}$$=\frac{72}{1}=72c{m}^{3}or0.072d{m}^3$