An alloy of metals X and Y weighs 12 g and c | vedclass

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(A) Mass of metal $X = 12 \ g 	imes 0.20 = 2.4 \ g$. Mass of metal $Y = 12 \ g - 2.4 \ g = 9.6 \ g$. Moles of $X = \frac{2.4 \ g}{40 \ g/mol} = 0.06

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(A) Mass of metal $X = 12 \ g 	imes 0.20 = 2.4 \ g$. Mass of metal $Y = 12 \ g - 2.4 \ g = 9.6 \ g$. Moles of $X = \frac{2.4 \ g}{40 \ g/mol} = 0.06 \ mol$. Since the ratio of atoms $X:Y$ is $2:5$,the ratio of moles $n_X:n_Y$ is also $2:5$. $n_Y = n_X 	imes \frac{5}{2} = 0.06 	imes 2.5 = 0.15 \ mol$. Atomic mass of $Y = \frac{	ext{Mass of } Y}{n_Y} = \frac{9.6 \ g}{0.15 \ mol} = 64 \ g/mol$. Thus,the atomic mass of $Y$ is $64 \ amu$.

An alloy of metals $X$ and $Y$ weighs $12 \ g$ and contains atoms $X$ and $Y$ in the ratio of $2:5$. The percentage of metal $X$ in the alloy is $20$ by mass. If the atomic mass of $X$ is $40$,what is the atomic mass of metal $Y$ (in $amu$)?

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