Question

What is the equivalent weight of methanol, if one mole of $CH _{3} OH$ 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 $CH _{3} OH$ :

$\underset{\text{Oxidation number of carbon in} CH_3OH}{CH _{3} OH + O _{2}} \longrightarrow \underset{\text{Oxidation number of carbon in} CO}{CO +2 H _{2} O}$

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

$x+3-2+1=0$

$x+2 =0$

$x =-2$

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

$x-2=0$

$x=+2$

Difference in oxidation number $=4$

Equivalent weight of methanol

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