When 1.164 g of a certain metal sulphide was roasted in air, 0.972 g of the metal oxide was formed. If the oxidation number of the metal is +2; calculate the molar mass(g) of the metal.
64

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Solution

The correct option is A 64
Sulphide is = MS (oxidation number of metal ion being +2) $\begin{matrix} M S (M + 32) 1.164 \end{matrix} + \frac{3}{2} O_{2} \to \begin{matrix} M O (M + 16) 0.972 \end{matrix} + S O_{2}$ $∴$ $\frac{M + 32}{1.164} = \frac{M + 16}{0.972}$ $∴$ $M = 64$

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When 1.164 g of a certain metal sulphide was roasted in air, 0.972 g of the metal oxide was formed. If the oxidation number of the metal is +2; calculate the molar mass(g) of the metal.64

The correct option is A 64Sulphide is = MS (oxidation number of metal ion being +2) MS(M+32)1.164+32O2→MO(M+16)0.972+SO2 ∴ M+321.164=M+160.972 ∴ M=64

When 1.164 g of a certain metal sulphide was roasted in air, 0.972 g of the metal oxide was formed. If the oxidation number of the metal is +2; calculate the molar mass(g) of the metal.
64

The correct option is A 64
Sulphide is = MS (oxidation number of metal ion being +2) $\begin{matrix} M S (M + 32) 1.164 \end{matrix} + \frac{3}{2} O_{2} \to \begin{matrix} M O (M + 16) 0.972 \end{matrix} + S O_{2}$ $∴$ $\frac{M + 32}{1.164} = \frac{M + 16}{0.972}$ $∴$ $M = 64$