Question

What is the equivalent weight of methanol, if one mole of $C{H}_{3}OH$ is combusted to form $CO$ and $H_{2}O$ ?

• A. 8
• B. 5.33
• C. 4
• D. 10.66

Answer 1

Answer: (A)

Combustion of one mole of $C{H}_{3}OH$ :

$\underset{\text{Oxidation number of carbon in}C{H}_{3}OH}{C{H}_{3}OH+O_2}⟶\underset{\text{Oxidation number of carbon in}CO}{CO+2H_{2}O}$

$x+(3\times 1)+(1\times -2)+(1\times 1)=0$

$x+3-2+1=0$

$x+2=0$

$x=-2$

$x+(1\times -2)=0$

$x-2=0$

$x=+2$

Difference in oxidation number $=4$

Equivalent weight of methanol

$=\frac{\text{ molecular weight of }C{H}_{3}OH}{\text{ change in oxidation number }}=\frac{32}{4}=8$