Question

If the weight of the metal chloride is x grams containing y grams of the metal, then what is the equivalent weight of the metal in terms of the x, y, and 35.5?

Answer 1

$C{{l}^{-}}$ is the anion (negatively charged ion) of the chloride. When the halogen chlorine acquires an electron, or when a molecule like hydrogen chloride is dissolved in water or other polar solvents, it is produced. Chloride salts, such as sodium chloride, are frequently extremely water soluble. It's an electrolyte that's found in all bodily fluids and is crucial for maintaining acid/base balance, sending nerve impulses, and controlling liquid movement in and out of the cells.

Complete answer:
The msinces of the one equivalent (also known since the grams equivalent) is the msinces of the a given material that will mix with or displace a fixed quantity of the another substance. The msinces of an element that combines with or displaces 1.008 grams of the hydrogen, 8.0 grams of the oxygen, or 35.5 grams of the chlorine is its equivalent weight. These figures are calculated by dividing the atomic weight by the typical valence.
The mole is the International System of the Units' basic unit of the material quantity (SI). It is defined as a collection of precisely $6.02214076\times {{10}^{23}}$ particles, which may be atoms, molecules, ions, or electrons.
The Avogadro number ($6.02214076\times {{10}^{23}}$) wsince set such that the msinces of the one mole of the a chemical compound in gram is numerically equivalent to the average msinces of the one molecule of the the compound in daltons for most practical applications.
Weight of the metal $=\mathbf{y} \mathbf{g}$
Weight of the metal chloride $=\mathbf{x} \mathbf{g}$
$\therefore$ Weight of the chlorine $=(\mathrm{x}-\mathbf{y}) \mathrm{g}$
Molecular weight of the chlorine $=\mathbf{3 5 . 5} \mathbf{g}$
Since we know that, No. of the moles $=\dfrac{\text { weight }}{\text { mol. wt. }}$
$\therefore$ No. of the moles of the chlorine $=\dfrac{(\mathbf{x}-\mathbf{y})}{\mathbf{3 5 . 5}}$
Let $E$ be the equivalent weight of the metal. Equivalent of the chlorine $=35.5 \mathrm{~g}$
Since we know that $\mathrm{gm}$ eq. $=\dfrac{\text { wt. }}{\text { eq. wt. }}$
$\therefore(\mathrm{gm} \text { eq. })_{\mathrm{Cl}}=(\mathrm{gm} \text { eq. })_{\mathrm{Metal}}$
$\Rightarrow \dfrac{(\mathrm{x}-\mathrm{y})}{35.5}=\dfrac{\mathrm{y}}{\mathbf{E}}$
$\Rightarrow \mathbf{E}=\dfrac{\mathbf{y}}{(\mathbf{x}-\mathbf{y})} \times \mathbf{3 5 . 5}$
Hence equivalent weight is $\Rightarrow \mathbf{E}=\dfrac{\mathbf{y}}{(\mathbf{x}-\mathbf{y})}\times \mathbf{35}.\mathbf{5}g$

Note:
In gravimetric analysis, the phrase "equivalent weight" meant the quantity of precipitate that corresponded to one grams of analyte (the species of interest). The various definitions arose from the habit of reporting gravimetric results as mass fractions of the analyte, which were frequently represented as percentages. The equivalency factor, which was the numerical factor by which the mass of precipitate had to be multiplied to acquire the amount of analyte, was a related word. It was one grams divided by equivalent weight.