
The largest number of hydrogen atoms present in
A.\[3.01 \times {10^{23}}\] molecules of \[N{H_3}\]
B.9.00 g of \[{H_2}O\]
C.0.4000 mol of \[C{H_4}\]
D.8.00 g of \[{N_2}{H_4}\]
Answer
486.9k+ views
Hint: According to the mole concept, 1 mole of any substance will contain the molar mass of that substance which in turn contains the Avogadro number of atoms. Hence we have to find the molar mass of each substance and thus find the molar mass of the substances and thus the number of hydrogen atoms in each of them.
Complete answer: Let,
For option A we are directly given the number of atoms and thus we can say that there is 0.5 moles of ammonia and thus there will be 3 times the number of hydrogen atoms in the given sample.
\[3 \times 3.01 \times {10^{23}}\] hydrogen atoms.
\[ \Rightarrow 9.03 \times {10^{23}}\] Hydrogen atoms
9.00 g of \[{H_2}O\] is also 0.5 moles of water. We know that one mole contains \[6.022 \times {10^{23}}\] molecules and therefore contains twice the amount of hydrogen atom. Thus 9 grams contain \[6.022 \times {10^{23}}\] hydrogen atoms.
\[\because 1mole \to 18g \to 6.022 \times {10^{23}}\]
\[ \Rightarrow 0.5mole \to 9g \to 3.022 \times {10^{23}}\] water molecules
One molecule of water has 2 hydrogen atoms and thus \[3.022 \times {10^{23}}\] molecules will contain \[6.022 \times {10^{23}}\] hydrogen atoms.
For (C) option we can say that:
\[\because 1mole \to 6.022 \times {10^{23}}\] molecules of methane
\[ \Rightarrow 0.4mole \to 2.4088 \times {10^{23}}\] molecules of methane
One molecule of methane contains 4 hydrogen atoms and thus \[2.4088 \times {10^{23}}\] molecules contain \[9.635 \times {10^{23}}\] atoms of hydrogen.
In option (D), we can do the following calculations:
\[\because 1mole \to 32g \to 6.022 \times {10^{23}}\]
\[ \Rightarrow 8g \to 1.5 \times {10^{23}}\] molecules of \[{N_2}{H_4}\]
Each molecule of \[{N_2}{H_4}\] contains 4 hydrogen. Thus \[1.5 \times {10^{23}}\] molecules of \[{N_2}{H_4}\] contains \[6.022 \times {10^{23}}\] atoms of hydrogen.
Thus we can say that 0.4 g of methane contains the largest number of hydrogen atoms.
Thus the correct option is (C).
Note:
While dealing with these kinds of questions we have to find out the number of atoms for each of the options. Also, remember to use the standard formulas for the number of moles which is given mass divided by molar mass. Another method is by using cross multiplication in a linear fashion. This is being used in the method above.
Complete answer: Let,
For option A we are directly given the number of atoms and thus we can say that there is 0.5 moles of ammonia and thus there will be 3 times the number of hydrogen atoms in the given sample.
\[3 \times 3.01 \times {10^{23}}\] hydrogen atoms.
\[ \Rightarrow 9.03 \times {10^{23}}\] Hydrogen atoms
9.00 g of \[{H_2}O\] is also 0.5 moles of water. We know that one mole contains \[6.022 \times {10^{23}}\] molecules and therefore contains twice the amount of hydrogen atom. Thus 9 grams contain \[6.022 \times {10^{23}}\] hydrogen atoms.
\[\because 1mole \to 18g \to 6.022 \times {10^{23}}\]
\[ \Rightarrow 0.5mole \to 9g \to 3.022 \times {10^{23}}\] water molecules
One molecule of water has 2 hydrogen atoms and thus \[3.022 \times {10^{23}}\] molecules will contain \[6.022 \times {10^{23}}\] hydrogen atoms.
For (C) option we can say that:
\[\because 1mole \to 6.022 \times {10^{23}}\] molecules of methane
\[ \Rightarrow 0.4mole \to 2.4088 \times {10^{23}}\] molecules of methane
One molecule of methane contains 4 hydrogen atoms and thus \[2.4088 \times {10^{23}}\] molecules contain \[9.635 \times {10^{23}}\] atoms of hydrogen.
In option (D), we can do the following calculations:
\[\because 1mole \to 32g \to 6.022 \times {10^{23}}\]
\[ \Rightarrow 8g \to 1.5 \times {10^{23}}\] molecules of \[{N_2}{H_4}\]
Each molecule of \[{N_2}{H_4}\] contains 4 hydrogen. Thus \[1.5 \times {10^{23}}\] molecules of \[{N_2}{H_4}\] contains \[6.022 \times {10^{23}}\] atoms of hydrogen.
Thus we can say that 0.4 g of methane contains the largest number of hydrogen atoms.
Thus the correct option is (C).
Note:
While dealing with these kinds of questions we have to find out the number of atoms for each of the options. Also, remember to use the standard formulas for the number of moles which is given mass divided by molar mass. Another method is by using cross multiplication in a linear fashion. This is being used in the method above.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

