1. Tardigrade
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  3. Chemistry
  4. 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

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Q. 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

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Solution:

$x$ is same in all and $y$ is changtng
in $A: B: C: D: E(Y$ amount $)=1: 2: 3: 4: 5$
$A$ has $28$ part of $X$ and $16$ part of $Y$so if A contain $16$ part of $y$ in $C$
$\frac{16 \times 3}{1} = 48$ part