Answer
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Hint: In order to deal with this question first we will define the atomic weight of the metal then we will proceed further by assuming the alkaline earth metal and atomic mass of metal as variables and according to it we will write the reaction and calculate the required atomic weight.
Complete step-by-step answer:
Formula used- $\dfrac{{{\text{Molecular mass of compound}}}}{{{\text{Atomic mass of metal}}}} = \dfrac{{{\text{weight of compound}}}}{{{\text{weight of metal}}}}$
Given that:
Weight of compound = 14.8 gm
Weight of metal = 12 gm
Let the mentioned alkaline earth metal in the problem be A
Now we will write the reaction:
$3A + {N_2} \to {A_3}{N_2}$
As we know the relation between the molar mass and the weight of metal in a compound is given as:
$ \Rightarrow \dfrac{{{\text{Molecular mass of compound}}}}{{{\text{Atomic mass of metal}}}} = \dfrac{{{\text{weight of compound}}}}{{{\text{weight of metal}}}}$
Let atomic mass of metal is $x$
Therefore molecular mass of the given compound is:
$ = 3x + 28$
By substituting the values in above formula we have
$
\because \dfrac{{{\text{Molecular mass of compound}}}}{{{\text{Atomic mass of metal}}}} = \dfrac{{{\text{weight of compound}}}}{{{\text{weight of metal}}}} \\
\Rightarrow \dfrac{{3x + 28}}{{3x}} = \dfrac{{14.8}}{{12}} \\
$
Let us solve the equation to find the value of x
$
\Rightarrow 12 \times \left( {3x + 28} \right) = 3x \times 14.8 \\
\Rightarrow 3x \times 14.8 - 12 \times 3x = 28 \times 12 \\
\Rightarrow 2.8 \times 3x = 28 \times 12 \\
\Rightarrow x = \dfrac{{28 \times 12}}{{2.8 \times 3}} \\
\Rightarrow x = 40 \\
$
Hence the atomic weight of metal is 40.
So, option C is the correct option.
Note- In group 2 of the periodic table the alkaline earth metals are six chemical elements. They consist of beryllium, magnesium, calcium, strontium, barium, radium. The elements have very similar properties: At normal temperature and pressure, they are all smooth, silvery-white, somewhat reactive metals. The name derives from the fact that, when dissolved in water, the oxides of these metals formed simple solutions, which stayed solid at the temperatures open to ancient alchemists. Like the Group 1A elements, the alkaline earth metals are too reactive in their elementary state to be present in nature.
Complete step-by-step answer:
Formula used- $\dfrac{{{\text{Molecular mass of compound}}}}{{{\text{Atomic mass of metal}}}} = \dfrac{{{\text{weight of compound}}}}{{{\text{weight of metal}}}}$
Given that:
Weight of compound = 14.8 gm
Weight of metal = 12 gm
Let the mentioned alkaline earth metal in the problem be A
Now we will write the reaction:
$3A + {N_2} \to {A_3}{N_2}$
As we know the relation between the molar mass and the weight of metal in a compound is given as:
$ \Rightarrow \dfrac{{{\text{Molecular mass of compound}}}}{{{\text{Atomic mass of metal}}}} = \dfrac{{{\text{weight of compound}}}}{{{\text{weight of metal}}}}$
Let atomic mass of metal is $x$
Therefore molecular mass of the given compound is:
$ = 3x + 28$
By substituting the values in above formula we have
$
\because \dfrac{{{\text{Molecular mass of compound}}}}{{{\text{Atomic mass of metal}}}} = \dfrac{{{\text{weight of compound}}}}{{{\text{weight of metal}}}} \\
\Rightarrow \dfrac{{3x + 28}}{{3x}} = \dfrac{{14.8}}{{12}} \\
$
Let us solve the equation to find the value of x
$
\Rightarrow 12 \times \left( {3x + 28} \right) = 3x \times 14.8 \\
\Rightarrow 3x \times 14.8 - 12 \times 3x = 28 \times 12 \\
\Rightarrow 2.8 \times 3x = 28 \times 12 \\
\Rightarrow x = \dfrac{{28 \times 12}}{{2.8 \times 3}} \\
\Rightarrow x = 40 \\
$
Hence the atomic weight of metal is 40.
So, option C is the correct option.
Note- In group 2 of the periodic table the alkaline earth metals are six chemical elements. They consist of beryllium, magnesium, calcium, strontium, barium, radium. The elements have very similar properties: At normal temperature and pressure, they are all smooth, silvery-white, somewhat reactive metals. The name derives from the fact that, when dissolved in water, the oxides of these metals formed simple solutions, which stayed solid at the temperatures open to ancient alchemists. Like the Group 1A elements, the alkaline earth metals are too reactive in their elementary state to be present in nature.
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