Question

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 $Y$
• B. $48$ parts by weight of $Y$
• C. $64$ parts by weight of $Y$
• D. $80$ parts by weight of $Y$

Answer 1

Answer: (B)

$x$ 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