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Standard XII

Chemistry

Laws of Definite Proportions, Multiple Proportions

Two elements,...

Question

Two elements, X (Atomic mass 16) and Y (Atomic mass 14) combine to form compounds A, B and C. The ratio of different masses of Y which combines with fixed mass of X in A, B, and C is $1 : 3 : 5$ . If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C 16 parts by mass of X will combine with

14 parts by mass of Y

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42 parts by mass of Y

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70 parts by mass of Y

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82 parts by mass of Y

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Solution

The correct option is C 70 parts by mass of Y
In compound B, 32 parts of X react with 84 parts of Y.
In compound C, 16 parts of X react with m parts of Y.
The ratio of masses of Y which combines with fixed mass of X in compounds B and C is $3 : 5$
$\begin{matrix} B & 84 / 32 & 3 C & m / 16 & 5 \end{matrix}$
$\Rightarrow \frac{84 / 32}{m / 16} = \frac{3}{5}$
$m = \frac{42 \times 5}{3} = 70$

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Laws of Definite Proportions, Multiple Proportions

Standard XII Chemistry

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The correct option is C 70 parts by mass of YIn compound B, 32 parts of X react with 84 parts of Y. In compound C, 16 parts of X react with m parts of Y. The ratio of masses of Y which combines with fixed mass of X in compounds B and C is 3:5 B84/323Cm/165 ⇒84/32m/16=35 m=42×53=70