4 , g of copper is dissolved in concentrated | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The equivalent weight of an element is defined as the mass of the element that combines with $8 \, g$ of oxygen.Given that $4 \, g$ of $Cu$ produc

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) The equivalent weight of an element is defined as the mass of the element that combines with $8 \, g$ of oxygen.Given that $4 \, g$ of $Cu$ produces $5 \, g$ of copper oxide.Therefore,$4 \, g$ of $Cu$ combines with $(5 - 4) = 1 \, g$ of oxygen.Using the unitary method:$1 \, g$ of oxygen combines with $4 \, g$ of $Cu$.So,$8 \, g$ of oxygen will combine with $\frac{4 	imes 8}{1} = 32 \, g$ of $Cu$.The equivalent weight of copper is $32 \, g \, eq^{-1}$.

$4 \, g$ of copper is dissolved in concentrated $HNO_3$. Upon strong heating of the resulting copper nitrate solution,$5 \, g$ of its oxide is obtained. The equivalent weight of copper is ...... .

Similar Questions

An element has two main isotopes of mass numbers $85$ and $87$. In nature,they occur in the ratio of $75\%$ and $25\%$ respectively. The atomic weight of the element will be approximately:

$1.520 \ g$ of the hydroxide of a metal on ignition gave $0.995 \ g$ of oxide. The equivalent weight of the metal is:

What is the equivalent mass of oxalic acid $(H_2C_2O_4 \cdot 2H_2O)$?

The equivalent mass of a divalent metal is $32$. What is the molar mass of its nitrate salt?

Which method is used to determine the molecular mass of a volatile substance?

Vedclass Test Series

Exam Paper Generator

Online Exam Module