MCQ
Metal oxide has $75\%$ metal. If vapour density of metal chloride is $89$ then atomic weight of metal will be
- A$59.5$
- ✓$72$
- C$24$
- D$48$
Molar mass of $M C l_{x}=2$ ltimes $89=178 \mathrm{g}$
$M+x \times 35.5 g / m o l=178$.......$(1)$
Given : metal oxide has $75 \%$ metal.
Thu formula of metal oxide $=M_{2} O_{x}$
percentage of metal $=\frac{2 M}{2 M+16 x}=0.75$
$0.5 \times M=12 x$
$M=24 x$......$(2)$
Solving $(1)$ and $(2)$
$x=3$
thus atomic mass of metal is $24 \times 3=72 g$
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$NH _{2} CN _{( s )}+\frac{3}{2} O _{2}( g ) \rightarrow N _{2( g )}+ O _{2}( g )+ H _{2} O _{(l)}$
is ............ $kJ$. (Rounded off to the nearest integer)
[Assume ideal gases and $\left. R =8.314\, J\, mol ^{-1} K ^{-1}\right]$