MCQ
Two elements $X$ and $Y$ have atomic weights of $14$ and $16$. They form a series of compounds $A, B, C, D$ and $E$ in which the same amount of element $X$, $Y $ is present in the ratio $1 : 2 : 3 : 4 : 5$. If the compound $A$ has $28$ parts by weight of $X $ and $16$ parts by weight of $Y$, then the compound of $C$ will have $28$ parts weight of $X$ and
  • A
    $32$ parts by weight of $Y$
  • $48$ parts by weight of $Y$
  • C
    $64$ parts by weight of $Y$
  • D
    $80$ parts by weight of $Y$

Answer

Correct option: B.
$48$ parts by weight of $Y$
b
The law of multiple proportions states that if two elements form more than one compound between them, then the ratios of the masses of the second element which combines with fixed masses of the first element will always be a ratio of a small whole number.

If compound $A$ has $28$ parts by weight of $X$ and $16$ parts by weight of $Y$, then fixing $X$, compound $C$ would have $3$ times as much $Y$ as compound $A=16 \times 3=48$ parts.

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