
What is the weight of bromine needed for the reaction with \[{\mathbf{21gm}}\] of \[{C_3}{H_6}\](atomic weight of $Br = 80)$
A.$40{\text{g}}$
B.$80{\text{g}}$
C.$160{\text{g}}$
D.$120{\text{g}}$
Answer
566.7k+ views
Hint:To answer this question, you should recall the concept of bromination of alkenes. Apply the mole concept in the reaction to find the answer to this question. Equal moles of bromine will react with the alkene.
Formula Used:${\text{moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
Complete step by step answer:
The reaction involving bromination of the aforementioned alkene can be represented by the equation: \[{C_3}{H_6} + B{r_2} \to {C_3}{H_6}B{r_2}\].
The molar mass of \[{C_3}{H_6} = 42{\text{gm}}\] and \[Br = 160{\text{ gm}}\]
Now from the reaction, we can conclude that \[42{\text{gm}}\] of \[{C_3}{H_6}\] needs \[160{\text{ gm}}\]of bromine.
Simply applying the unitary method: \[1{\text{gm}}\] of \[{C_3}{H_6}\]would need \[160/42{\text{g}}\] of bromine.
$\therefore $\[21{\text{ gm}}\]of \[{C_3}{H_6}\] would require \[\dfrac{{160}}{{42}} \times 21 = 80{\text{gm}}\;\]of bromine.
Thus, the correct option is B.
Note:
The concept that a mole of any substance contains the same number of particles was formed out of research which was conducted by Italian physicist Amedeo Avogadro. Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: $6.022 \times {10^{23}}$ per mol.
It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$ in 12 grams of that substance and is the reciprocal of atomic mass in grams. Now the mole concept can be applied to ions and formula units.
Illustration 1: 1 mole of \[{O_2}\] means Avogadro's number of oxygen molecules and it will be equal to 2 times Avogadro's number of oxygen atoms.
Illustration 2: 1 g-molecule of \[{O_2}\] is the same as one mole of the oxygen molecule and contains Avogadro's number of nitrogen molecules and twice this number of atoms.
Illustration 3: 1 g-atom of Nitrogen means 1 mole of nitrogen atoms and contains Avogadro's number of nitrogen atoms. The mass of 1 mole of any species is equal to its molar mass.
The formulae for the mole concept can be summarized as:
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}.
Formula Used:${\text{moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$
Complete step by step answer:
The reaction involving bromination of the aforementioned alkene can be represented by the equation: \[{C_3}{H_6} + B{r_2} \to {C_3}{H_6}B{r_2}\].
The molar mass of \[{C_3}{H_6} = 42{\text{gm}}\] and \[Br = 160{\text{ gm}}\]
Now from the reaction, we can conclude that \[42{\text{gm}}\] of \[{C_3}{H_6}\] needs \[160{\text{ gm}}\]of bromine.
Simply applying the unitary method: \[1{\text{gm}}\] of \[{C_3}{H_6}\]would need \[160/42{\text{g}}\] of bromine.
$\therefore $\[21{\text{ gm}}\]of \[{C_3}{H_6}\] would require \[\dfrac{{160}}{{42}} \times 21 = 80{\text{gm}}\;\]of bromine.
Thus, the correct option is B.
Note:
The concept that a mole of any substance contains the same number of particles was formed out of research which was conducted by Italian physicist Amedeo Avogadro. Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: $6.022 \times {10^{23}}$ per mol.
It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$ in 12 grams of that substance and is the reciprocal of atomic mass in grams. Now the mole concept can be applied to ions and formula units.
Illustration 1: 1 mole of \[{O_2}\] means Avogadro's number of oxygen molecules and it will be equal to 2 times Avogadro's number of oxygen atoms.
Illustration 2: 1 g-molecule of \[{O_2}\] is the same as one mole of the oxygen molecule and contains Avogadro's number of nitrogen molecules and twice this number of atoms.
Illustration 3: 1 g-atom of Nitrogen means 1 mole of nitrogen atoms and contains Avogadro's number of nitrogen atoms. The mass of 1 mole of any species is equal to its molar mass.
The formulae for the mole concept can be summarized as:
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

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

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

