Q2 / 125 · BITSAT 2011 · Chemistry

$1.520 \mathrm{~g}$ of hydroxide of a metal on ignition gave $0.995 \mathrm{~g}$ of oxide. The equivalent weight of metal is:

  • A
    1.52
  • B
    0.995
  • C
    190
  • Correct answer
    D
    9

Solution

Let $E$ be the equivalent weight of the metal $$ \text { So, } \frac{E+17}{E+8}=\frac{1.52}{0.995} $$ [17 is equivalent weight of $\mathrm{OH}$ and 8 is equivalent weight of oxygen] $$ \begin{array}{l} \Rightarrow 0.995 \mathrm{E}+17 \times 0.995=\mathrm{E} \times 1.52+8 \times 1.52 \\ \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ \therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9 \end{array} $$

Hence, the correct answer is option: 'D'.

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