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Match the following, and select the correct code. (Atomic mass of Chlorine = 35.5)
Column-I Column -II
A) Same number of atoms P) 4.25 g $N H_{3}$ and 4.5 g of $H_{2} O$
B) Same number of molecules Q) 2.20 g $C O_{2}$ and 0.90 g of $H_{2} O$
C) Same numbers of atoms as well as molecules R) 4.0 g $C H_{3} C l$ and 5.0 g of $N H_{3}$
D) Different numbers of atoms as well as molecules S) 3.2 g $O_{2}$ and 2.80 g of $C O$

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$P - D, Q - A, R - B, S - C$

$P - A, Q - B, R - C, S - D$

$P - B, Q - C, R - D, S - A$

$P - D, Q - B, R - C, S - A$

answer is B.

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Detailed Solution

In 4.25g $N H_{3}$ No of moles $= \frac{4.25}{17} = \frac{1}{4}$ No of molecules $= \frac{1}{4} N_{A}$ No of atoms $= \frac{1}{4} N_{A} \times 4 = N_{A}$	In 4.5 g $H_{2} O$ No of moles $= \frac{4.5}{18}$ No of molecules $= \frac{1}{4} N_{A}$ No of atoms $= \frac{3}{4} N_{A}$
In 2.2 g $C O_{2}$ No of moles $= \frac{2.2}{44} = \frac{1}{20}$ No of molecules $= \frac{1}{20} N_{A}$ No of atoms $= \frac{3}{20} N_{A}$	In 0.9 g $H_{2} O$ No of moles $= \frac{0.9}{18} = \frac{1}{20}$ No of molecules $= \frac{1}{20} N_{A}$ No of atoms $= \frac{3}{20} N_{A}$
In 3.2 g $O_{2}$ No moles $= \frac{3.2}{44} = 0.1$ No of molecules $= 0.1 N_{A}$ No of atoms $= 0.2 N_{A}$	In 2.8 g CO No of moles $= \frac{2.8}{28} = 0.1$ No of molecules $= 0.1 N_{A}$ No of atoms $= 0.2 N_{A}$

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