Equivalent weights of X_2Y and X_2Y_3 are 38 | vedclass

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(B) The equivalent weight $E$ is defined as the molar mass divided by the total valence change or total charge.For $X_2Y$,the valency of $Y$ is $2$ an

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$30, 16$

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(B) The equivalent weight $E$ is defined as the molar mass divided by the total valence change or total charge.For $X_2Y$,the valency of $Y$ is $2$ and $X$ is $1$. The equivalent weight is $E_{X_2Y} = \frac{M_{X_2Y}}{n-factor}$.Assuming $X$ has valency $1$ and $Y$ has valency $2$,$E_{X_2Y} = \frac{2X + Y}{2} = 38 \Rightarrow 2X + Y = 76$ (Equation $1$).For $X_2Y_3$,the valency of $Y$ is $2$ and $X$ is $3$. The equivalent weight is $E_{X_2Y_3} = \frac{2X + 3Y}{6} = 18 \Rightarrow 2X + 3Y = 108$ (Equation $2$).Subtracting Equation $1$ from Equation $2$:$(2X + 3Y) - (2X + Y) = 108 - 76$$2Y = 32 \Rightarrow Y = 16$.Substituting $Y = 16$ into Equation $1$:$2X + 16 = 76$ $\Rightarrow 2X = 60$ $\Rightarrow X = 30$.Thus,the atomic masses are $X = 30$ and $Y = 16$.

Equivalent weights of $X_2Y$ and $X_2Y_3$ are $38$ and $18$ respectively. Find the atomic masses of $X$ and $Y$:-

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