Answer
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Hint: In order to deal with this question we will first define the term equivalent mass further we will evaluate charge of nitrogen and at last by putting the values of molar mass of nitrogen and charge of nitrogen in formula we will get the required answer.
Formula used- ${\text{Equivalent mass of nitrogen}} = \dfrac{{{\text{Molar mass of nitrogen}}}}{{{\text{Charge of nitrogen in }}{N_2}{O_5}}}$
Complete step by step answer:
Equivalent mass of an atom in a given compound is defined as the ratio of its molecular mass and the charge of an atom in that compound.
We know that the chemical formula for nitrogen pentoxide is ${N_2}{O_5}$ .
As we know that the equivalent mass of the atom depends on both the molar mass and the charge on the atom. So let us find out both the molar mass of nitrogen and the charge on nitrogen in the given compound.
First we will evaluate the charge of nitrogen
In the given compound having chemical formula ${N_2}{O_5}$ , let x be the charge on nitrogen in the compound. Also we know that the charge on the oxygen is -2. So let us solve for x.
$
\Rightarrow 2x + 5\left( { - 2} \right) = 0 \\
\Rightarrow 2x - 10 = 0 \\
\Rightarrow 2x = 10 \\
\Rightarrow x = + 5 \\
$
As we know that, the molar mass of nitrogen is 14 u as the atomic number of nitrogen in the modern periodic table is 7.
Now we have both the molar mass and the charge of nitrogen.
As we know the equivalent mass of nitrogen is given as:
${\text{Equivalent mass of nitrogen}} = \dfrac{{{\text{Molar mass of nitrogen}}}}{{{\text{Charge of nitrogen in }}{N_2}{O_5}}}$
Putting values in above equation, we get
$
\Rightarrow {\text{Equivalent mass of nitrogen}} = \dfrac{{14}}{5} \\
= 2.8g/eq. \\
$
Hence, the equivalent mass of nitrogen in the given compound is $2.8g/eq.$
So, the correct answer is “Option C”.
Note: In order to solve such types of problems students must remember the formula for the calculation of equivalent mass of the atom from the molar mass of the atom. Also the students must remember the method to find out the charge on the atom in a compound. Also the students must know the method to find the molar mass from the periodic table on the basis of atomic number.
Formula used- ${\text{Equivalent mass of nitrogen}} = \dfrac{{{\text{Molar mass of nitrogen}}}}{{{\text{Charge of nitrogen in }}{N_2}{O_5}}}$
Complete step by step answer:
Equivalent mass of an atom in a given compound is defined as the ratio of its molecular mass and the charge of an atom in that compound.
We know that the chemical formula for nitrogen pentoxide is ${N_2}{O_5}$ .
As we know that the equivalent mass of the atom depends on both the molar mass and the charge on the atom. So let us find out both the molar mass of nitrogen and the charge on nitrogen in the given compound.
First we will evaluate the charge of nitrogen
In the given compound having chemical formula ${N_2}{O_5}$ , let x be the charge on nitrogen in the compound. Also we know that the charge on the oxygen is -2. So let us solve for x.
$
\Rightarrow 2x + 5\left( { - 2} \right) = 0 \\
\Rightarrow 2x - 10 = 0 \\
\Rightarrow 2x = 10 \\
\Rightarrow x = + 5 \\
$
As we know that, the molar mass of nitrogen is 14 u as the atomic number of nitrogen in the modern periodic table is 7.
Now we have both the molar mass and the charge of nitrogen.
As we know the equivalent mass of nitrogen is given as:
${\text{Equivalent mass of nitrogen}} = \dfrac{{{\text{Molar mass of nitrogen}}}}{{{\text{Charge of nitrogen in }}{N_2}{O_5}}}$
Putting values in above equation, we get
$
\Rightarrow {\text{Equivalent mass of nitrogen}} = \dfrac{{14}}{5} \\
= 2.8g/eq. \\
$
Hence, the equivalent mass of nitrogen in the given compound is $2.8g/eq.$
So, the correct answer is “Option C”.
Note: In order to solve such types of problems students must remember the formula for the calculation of equivalent mass of the atom from the molar mass of the atom. Also the students must remember the method to find out the charge on the atom in a compound. Also the students must know the method to find the molar mass from the periodic table on the basis of atomic number.
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