Correct Answer - Option 1 : 27
Concept:
Equivalent weight/ Equivalent mass :
Equivalent weight = Molecular weight (or) Atomic weight / Valency factor
Equivalent weight = Molecular weight / n-factor
\(Eq.\,wt\, = \,\frac{{\text{M}}}{x}\)
Where n-factor varies for different species as follows:
1. For elements: x = valency of element
2. For acid: x = no. of replaceable H+ ions per acid molecule
3. For the base: x = no. of replaceable OH- ions per basic molecule
Calculation:
Given: Formula of compound= MCl3, Percentage of metal (M) present= 20%
To Find: Atomic weight of metal=?
Assume that 100 g of compound MCl3 contains 20 g of metal (M) and 80 g of Cl.
\(\frac{{{E_M}}}{{{w_M}}} = \frac{{{E_{Cl}}}}{{{w_{Cl}}}}\) ......(1)
Where, EM & ECl= eq. wt of metal and chlorine (35.5) respectively, wM & wCl= mass of metal (20g) and Cl (80g) present in MCl3
On substituting the values in Eqn (1)
\(\frac{{{E_M}}}{{20}} = \frac{{35.5}}{{80}}\)
\({E_M} = \frac{{35.5 \times 20}}{{80}} = 8.875\)
Therefore, the equivalent weight of metal = 8.875
Valency of metal in MCl3 = 3
Valency of Cl in MCl3 = 1
\({\text{Eq}}{\text{.}}\,{\text{wt}}\,{\text{of}}\,{\text{metal}}\,{\text{ = }}\,\frac{{{\text{Atomic}}\,{\text{weight}}}}{{{\text{valency}}\,{\text{of}}\,{\text{metal}}}}\)
\(8.875\,{\text{ = }}\,\frac{{{\text{Atomic}}\,{\text{weight}}}}{3}\)
So, the Atomic weight of metal is
= 8.875 × 3
= 26.625 g ≈ 27 g
Hence, the Atomic weight of Metal (M) in MCl3 is 27 g.