Answer
424.2k+ views
Hint: The empirical formula of the compound is given. If we can find out the percentage weight of the metal, we can find out the atomic weight by using the atomic weight of another component (nitrogen) of the metal nitride.
Complete step by step answer:
Let us consider, the mass of the compound is 100g.
According to the question, the compound contains 28% nitrogen by weight I.e. It contains 28g of nitrogen.
Therefore, the metal content is (100-28) = 72g I.e. in 100g of compound, 72g metal is present and 28g nitrogen is present.
As given in the question, the empirical formula of the compound is ${{M}_{3}}{{N}_{2}}$.
From the formula it can be said that it contains 2 moles of nitrogen and 3 moles of the metal ‘M’.
As we know, the sum of atomic weight of a molecule and the number of moles of molecule present gives us the total mass of the compound.
Therefore, (number of moles of metal$\times $atomic weight of metal) + (number of moles of nitrogen$\times $ atomic weight of nitrogen) = total mass of the metal nitride compound.
To solve this, let us consider, the atomic weight of the metal is ‘x’.
$\therefore 3x+(14\times 2)=100$
Or, $x=\dfrac{(100-28)}{3}$
Or, x= 24
Therefore, the atomic weight of the metal is 24.
Therefore the correct option is [D] 24.
Note:
The same problem can also be solved without considering the total mass of the compound. The formula,$\dfrac{weight}{no.of moles}$ will also give the atomic weight of the required metal.
The weight of the metal in the compound will be 72g.
$\dfrac{72}{3}$=24, which is equal to our calculated value above.
Complete step by step answer:
Let us consider, the mass of the compound is 100g.
According to the question, the compound contains 28% nitrogen by weight I.e. It contains 28g of nitrogen.
Therefore, the metal content is (100-28) = 72g I.e. in 100g of compound, 72g metal is present and 28g nitrogen is present.
As given in the question, the empirical formula of the compound is ${{M}_{3}}{{N}_{2}}$.
From the formula it can be said that it contains 2 moles of nitrogen and 3 moles of the metal ‘M’.
As we know, the sum of atomic weight of a molecule and the number of moles of molecule present gives us the total mass of the compound.
Therefore, (number of moles of metal$\times $atomic weight of metal) + (number of moles of nitrogen$\times $ atomic weight of nitrogen) = total mass of the metal nitride compound.
To solve this, let us consider, the atomic weight of the metal is ‘x’.
$\therefore 3x+(14\times 2)=100$
Or, $x=\dfrac{(100-28)}{3}$
Or, x= 24
Therefore, the atomic weight of the metal is 24.
Therefore the correct option is [D] 24.
Note:
The same problem can also be solved without considering the total mass of the compound. The formula,$\dfrac{weight}{no.of moles}$ will also give the atomic weight of the required metal.
The weight of the metal in the compound will be 72g.
$\dfrac{72}{3}$=24, which is equal to our calculated value above.
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