When 1.0 g of a metal nitrate is converted i | vedclass

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(D) Let the equivalent weight of the metal be $E$. The equivalent weight of the nitrate ion $(NO_3^-)$ is $62/1 = 62$. The equivalent weight of the su

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$38$

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(D) Let the equivalent weight of the metal be $E$. The equivalent weight of the nitrate ion $(NO_3^-)$ is $62/1 = 62$. The equivalent weight of the sulfate ion $(SO_4^{2-})$ is $96/2 = 48$. According to the law of equivalence,the number of equivalents of metal nitrate equals the number of equivalents of metal sulfate. $\frac{	ext{Mass of metal nitrate}}{	ext{Equivalent weight of metal nitrate}} = \frac{	ext{Mass of metal sulfate}}{	ext{Equivalent weight of metal sulfate}}$ $\frac{1.0}{E + 62} = \frac{0.86}{E + 48}$ $1.0(E + 48) = 0.86(E + 62)$ $E + 48 = 0.86E + 53.32$ $E - 0.86E = 53.32 - 48$ $0.14E = 5.32$ $E = \frac{5.32}{0.14} = 38$ Thus,the equivalent weight of the metal is $38$.

When $1.0 \ g$ of a metal nitrate is converted into $0.86 \ g$ of metal sulfate,what is the equivalent weight of the metal?

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