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Q.
$1.520 \,g$ of hydroxide of a metal on ignition gave $0.995\, g$ of oxide. The equivalent weight of metal is
UP CPMTUP CPMT 2006
Solution:
$ \frac{\text{Weight of metal hydroxide}}{\text{Weight of metal oxide}}$
$=\frac{\text{Equivalent weight of metal hydroxide}}{\text{Equivalent weight of metal oxide}}$
or, $ \frac{1.520}{0.995}=\frac{E_{m}+17}{E_{m}+8}$
[Where $E_{m}$ =Equivalent weight of metal ]
or, $1.528=\frac{E_{m}+17}{E_{m}+8}$
or, $0.528 E_{m}=4.78$
or, $E_{m}=9$