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  4. 1.520 g of the hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent weight of metal is

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Q. $1.520\, g$ of the hydroxide of a metal on ignition gave $0.995\, g$ of oxide. The equivalent weight of metal is

Uttarkhand PMTUttarkhand PMT 2007

Solution:

$\frac{\text { Wt. of metal hydroxide }}{\text { Wt. of metal oxide }}$
$=\frac{\text { Eq. wt. of metal }+\text { Eq. wt. of } OH ^{-}}{\text {Eq. wt. of metal }+\text { Eq. wt. } O^{2-}}$
$=\frac{1.520}{0.995}=\frac{E+17}{E+8}$
$1.520(E+8)=0.995(E+17)$
$1.520 E+12.160=0.995 E+16.915$
$1.520 E-0.995 E=16.915-12.16$
or $0.525 E=16.915-12.16=4.755$
$E=\frac{4.755}{0.525}=9.0$