Question

The number of molecules present in $ 1.42g $ of chlorine are: $ \left( {{N_A} = 6 \times {{10}^{23}};Cl = 35.5} \right) $
(A) $ 6 \times {10^{21}} $
(B) $ 6 \times {10^{22}} $
(C) $ 1.2 \times {10^{22}} $
(D) $ 1.2 \times {10^{21}} $

Answer 1

Hint: Number of molecules is defined as the product of the number of moles and Avogadro’s number. The value of Avogadro’s number is $ 6.022 \times {10^{23}} $ . The number of moles can be defined as the ratio of the given mass of a substance and the molecular mass of that substance.

Complete answer:
Number of molecules = Number of moles $ \times $ $ {N_A} $ $ \to 1 $
Where, $ {N_A} $ is Avogadro's number. Firstly, we calculate the number of moles of a substance.
Number of moles $ = \dfrac{{Given{\text{ mass}}}}{{Molecular{\text{ mass}}}} $
Let the number of moles be n, given mass be m and molecular mass be M.
So, $ n = \dfrac{m}{M} $
 $ m = 1.42g $ and $ M = 71g $
Molecular mass of chlorine is $ 71 $ because it exists in $ C{l_2} $ form and the mass of one $ Cl $ is $ 35.5 $ . Therefore, the mass of two $ Cl $ atoms is $ 71 $ .
Therefore, $ n = \dfrac{{1.42}}{{71}} $
 $ n = 0.02 $
Put this value of n in $ 1 $ we get,
Number of molecules $ = 0.02 \times 6.022 \times {10^{23}} $
 $ = 0.120 \times {10^{23}} $
 $ = $ $ 1.2 \times {10^{22}} $
Therefore, the number of molecules present in $ 1.42g $ of chlorine are $ 1.2 \times {10^{22}} $ .
Therefore, option C is the correct option.

Note:
Chlorine is the element of group $ 17 $ and its atomic number is also $ 17 $ . It belongs to the halogen group. It is heavier than fluorine $ \left( F \right) $ but lighter than bromine $ \left( {Br} \right) $ and iodine $ \left( I \right) $ . At room temperature, it is present as yellow-green gas. Chlorine is a strong oxidizing agent and also an extremely reactive element. Chlorine has the highest electron affinity. Its electronegativity is also very high. It lags behind only from oxygen and fluorine in terms of electronegativity. Chlorine is also used in the manufacture of common salt $ \left( {NaCl} \right) $ .