JEE MAIN - Chemistry (2018 - 16th April Morning Slot - No. 11)
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 $$ \times $$ 1023 mol-1)
(Atomic wt. of Cl = 35.5 u; Avogadro constant = 6.023 $$ \times $$ 1023 mol-1)
6.023 $$ \times $$ 1020
6.023 $$ \times $$ 109
6.023 $$ \times $$ 1021
6.023 $$ \times $$ 1023
Explanation
% of Cl = 3.55
$$\therefore\,\,\,\,$$ In 100 g chlorohydrocarbon 3.55 gm Cl present.
In 1 gm chlorohydrocarbon Cl present
= $${{3.55} \over {100}}$$
= 0.0355 gm
$$\therefore\,\,\,\,$$ No of Moles of Cl = $${{0.0355} \over {35.5}}$$
= 0.001 mole
$$\therefore\,\,\,\,$$ no of Cl atoms = 0.001 $$ \times $$ 6.023 $$ \times $$ 103
= 6.023 $$ \times $$ 1020
$$\therefore\,\,\,\,$$ In 100 g chlorohydrocarbon 3.55 gm Cl present.
In 1 gm chlorohydrocarbon Cl present
= $${{3.55} \over {100}}$$
= 0.0355 gm
$$\therefore\,\,\,\,$$ No of Moles of Cl = $${{0.0355} \over {35.5}}$$
= 0.001 mole
$$\therefore\,\,\,\,$$ no of Cl atoms = 0.001 $$ \times $$ 6.023 $$ \times $$ 103
= 6.023 $$ \times $$ 1020
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