An unknown chlorohydrocarbon has 3.55 % of chlorine. If each...

An unknown chlorohydrocarbon has 3.55 % of chlorine. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are :(Atomic wt. of

Cl = 35 .5 u;

Avogadro constant

= 6 .023 \times {10}^{23}{mol}^{– 1}

)

Options

A $6.023 \times {10}^{9}$
B $6.023 \times {10}^{23}$
C $6.023 \times {10}^{21}$
D $6.023 \times {10}^{20}$

Correct answer

D. $6.023 \times {10}^{20}$

Step-by-step solution

{C}_{x}{H}_{y}Cl

% Cl = 3 .55

Weight of

Cl = 1 \times \dfrac{3.55}{100}

{n}_{C{l}^{-}} = \dfrac{1 \times 3.55}{100 \times 35.5}

No . of{Cl}^{-}ion = \dfrac{1 \times 3 .55}{100 \times 35 .5} \times 6 .023 \times {10}^{23} = 6 .023 \times {10}^{20}

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