What is the equivalent weight of methanol if one mole of `CH_3 OH` is combusted to form CO and `H_2 O` ?

Options

A. 8
B. 5.33
C. 4
D. 10.66

Answer

A

Explanation

Combustion of one mole of `CH_3OH`:
`{:(CH_3OH+O_2rarr),("Oxidation nuber of carbon"),("in "CH_3OH):}{:(CO+2H_2O),("Oxidation number of"),("carbon in CO"):}`
`x+(3xx1)+(1xx-2)+(1xx1)=0" "x+(1xx-2)=0`
`x+3-2+1=0" "x-2=0`
`x+2=0" "x=+2`
`x = -2`
Difference in oxidation number = 4
Equivalent weight of methanol
`=("molecular weight of "CH_3OH)/("change in oxidation number")=32/4=8`