1.5 gram of a divalent metal displace 4 gram of copper from a solution of copper sulphate the atomic weight of the metal is--

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Solution

Dear Student,

$\frac{Weight of metal}{Weight of Copper} = \frac{Eq . wt of metal}{Eq . wt of copper} Eq . wt of metal = \frac{Weight of metal}{Weight of Copper} \times Eq . wt of copper = \frac{1.5}{4} \times 32 = 12 Hence, atomic weight of metal = 12 \times 2 = 24$

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1.5 gram of a divalent metal displace 4 gram of copper from a solution of copper sulphate the atomic weight of the metal is--

Dear Student, Weight of metalWeight of Copper=Eq. wt of metalEq. wt of copperEq. wt of metal = Weight of metalWeight of Copper×Eq. wt of copper =1.54×32 =12Hence, atomic weight of metal = 12×2=24

Dear Student,

$\frac{Weight of metal}{Weight of Copper} = \frac{Eq . wt of metal}{Eq . wt of copper} Eq . wt of metal = \frac{Weight of metal}{Weight of Copper} \times Eq . wt of copper = \frac{1.5}{4} \times 32 = 12 Hence, atomic weight of metal = 12 \times 2 = 24$