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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Chemistry
Chapter
Redox Reactions

BITSAT 2011 · Chemistry · Redox Reactions

1.520 g of hydroxide of a metal on ignition gave 0.995 g of oxide....

1.520g1.520 \mathrm{ g} of hydroxide of a metal on ignition gave 0.995g0.995 \mathrm{ g} of oxide. The equivalent weight of metal is:

Options

  1. A1.52
  2. B0.995
  3. C190
  4. D9

Correct answer

D. 9

Step-by-step solution

Let EE be the equivalent weight of the metal  So, E+17E+8=1.520.995[17isequivalentweightofOHand8isequivalentweightofoxygen]\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]⇒0.995E+17×0.995=E×1.52+8×1.52⇒0.525E=16.915−12.16=4.755∴E=4.7550.525=9\begin{array}{l} \Rightarrow0.995 \mathrm{E}+17 \times0.995=\mathrm{E} \times1.52+8 \times1.52 \\ \Rightarrow 0.525 \mathrm{E}=16.915-12.16=4.755 \\ \therefore \quad \mathrm{E}=\frac{4.755}{0.525}=9 \end{array}$$
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