Question

The equivalent weight of $CaC{l_2}$ is:
A. $\dfrac{{Formula\;weight}}{2}$
B. $\dfrac{{Formula\;weight}}{1}$
C. $\dfrac{{Formula\;weight}}{3}$
D. $\dfrac{{Formula\;weight}}{4}$

Answer 1

Hint: The equivalent weight is determined by dividing the atomic weight or molecular weight or formula weight of the compound with the valency of the atom. The valency is defined as the combining capacity of the atom.

Complete answer:Given,
The equivalent weight is the mass in grams which is chemically the same as eight grams of oxygen or one gram of hydrogen. The equivalent weight is equal to the formula weight or atomic weight or molecular weight divided by the valency of the atom.
The formula for calculating the equivalent weight is shown below.
$E.W = \dfrac{F}{V}$
Where,
E.W is the equivalent weight.
F is the formula weight of the compound.
V is the valency.
The given compound is calcium chloride $CaC{l_2}$ where one calcium atom and two chlorine atoms are present.
The formula weight of calcium chloride is calculated as the sum of atomic masses of individual atoms multiplied with the number of atoms.
The atomic number of calcium is 40, the atomic number of chlorine is 35.5.
The formula weight of calcium chloride is calculated as shown below.
$F = 40 + 2 \times 35.5$
$\Rightarrow F = 111g/mol$
The charge on calcium cation is +2.
To calculate the equivalent weight of calcium chloride, substitute the value of formula weight and valency in the equation.
$\Rightarrow Eq.W = \dfrac{{111g/mol}}{2}$
$\Rightarrow Eq.W = 55.5$
Thus, the equivalent weight of $CaC{l_2}$ is given as shown below.
$Eq.W = \dfrac{{Formula\;weight}}{2}$
Therefore, the correct option is A.

Note:The formula weight is the other name for molecular weight. The electronic configuration of calcium is $[Ar]4{s^2}$, it can lose its two electrons to form a stable configuration.