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BITSAT 2011 Paper

Section: Chemistry

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Question : 42 of 150

Marks: +1, -0

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

1.52
0.995
190
9

Solution:

Let E be the equivalent weight of the metal

So ,

\frac{E+17}{E+8}

=

\frac{.52}{0.995}

[17 is equivalent weight of OH and 8 is equivalent weight of oxygen]

⇒ 0.995 E + 17 × 0.995 = E × 1.52 + 8 × 1.52

⇒ 0.525 E = 16. 915 – 12.16 = 4.755

∴ E =

\frac{4.755}{0.525}

= 9

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