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Standard XII

Chemistry

Laws of Definite Proportions, Multiple Proportions

X and Y are t...

Question

X and Y are two elements which can form $X_{2} Y$ and $X_{2} Y_{3}$ . 0.2 mole $X_{2} Y_{3}$ weighs 15.2 g and 0.5 mole $X_{2} Y$ weighs 22 g. Atomic weights of X and Y are:

20, 15

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31, 16

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14, 16

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16, 27

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Solution

The correct option is B $14, 16$
Let the atomic mass of X and Y be $M_{x} a n d M_{y}$
According to question,
$0.2 \times [2 M_{x} + 3 M_{y}] = 15.2 g$
and
$0.5 \times [2 M_{x} + M_{y}] = 22 g$
Solving the two equations we get, $M_{x} = 14$ g and $M_{y} = 16$ g

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X and Y are two elements which can form X2Y and X2Y3. 0.2 mole X2Y3 weighs 15.2 g and 0.5 mole X2Y weighs 22 g. Atomic weights of X and Y are:

The correct option is B 14,16Let the atomic mass of X and Y be MxandMyAccording to question,0.2×[2Mx+3My]=15.2gand 0.5×[2Mx+My]=22gSolving the two equations we get, Mx=14 g and My=16 g

X and Y are two elements which can form $X_{2} Y$ and $X_{2} Y_{3}$ . 0.2 mole $X_{2} Y_{3}$ weighs 15.2 g and 0.5 mole $X_{2} Y$ weighs 22 g. Atomic weights of X and Y are:

The correct option is B $14, 16$
Let the atomic mass of X and Y be $M_{x} a n d M_{y}$
According to question,
$0.2 \times [2 M_{x} + 3 M_{y}] = 15.2 g$
and
$0.5 \times [2 M_{x} + M_{y}] = 22 g$
Solving the two equations we get, $M_{x} = 14$ g and $M_{y} = 16$ g