MCQ
The oxide of a metal has $32\%$ oxygen. Its equivalent weight would be
- A$34$
- B$32$
- ✓$17$
- D$8$
✓
Answer
Correct option: C.
$17$
c
(c) Let weight of metal oxide = $100\,gm$
(c) Let weight of metal oxide = $100\,gm$
Weight of oxygen = $32\,gm$
$\therefore $ weight of metal $ = 100 - 32 = 68\,gm$
Equivalent weight of oxide $ = \frac{{{\rm{wt}}{\rm{. of metal}}}}{{{\rm{wt}}{\rm{. of oxygen}}}} \times 8$ $ = \frac{{68}}{{32}} \times 8 = 17$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The pair of species with similar shape is
In the complete combustion of $C_4H_{10}$ the number of oxygen moles required is
→Rate of hydration of following will be in order :
Which of the following is most reactive towards nucleophilic attack
The equilibrium constant for equilibrium is $2H{X_{(g)}}\,\, \rightleftharpoons \,{H_{2(g)}}\, + \,{X_{2(g)}}$ is $1\, \times \,10^{-5}$. What will be the equilibrium concentration of $HX$ if the equilibrium concentrations for $H_2$ and $X_2$ are $1.2\, \times \,10^{-3}\, M$ and $1.2 \times \, 10^{-4}\, M$ respectively
→Oxidation number of $N$ in ${(N{H_4})_2}S{O_4}$ is
→Arrange following anions in decresing order of their stability
→$\mathop C\limits^\Theta {H_3},\mathop N\limits^\Theta {H_2},\,\mathop O\limits^\Theta H,\,\mathop C\limits^\Theta l$
If equilibrium constant for reaction $2AB$ $\rightleftharpoons$ ${A_2} + {B_2}$, is $49$, then the equilibrium constant for reaction $AB$ $\rightleftharpoons$ $\frac{1}{2}{A_2} + \frac{1}{2}{B_2}$, will be
→An ideal gas expands in volume from $1 \times {10^{ - 3}}\,{m^3}$ to $1 \times {10^{ - 2}}\,{m^3}$ at $ 300 \,K $ against a constant pressure of $1 \times {10^5}\,N{m^{ - 2}}$. The work done is
→Which behaves both as a nucleophile and electrophile