Two bulbs B_1 and B_2 contain 10 g each of O | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For bulb $B_1$ ($10 \ g$ of $O_2$):Molar mass of $O_2 = 32 \ g/mol$.Number of molecules $= (10 / 32) 	imes N_A = 0.3125 \ N_A$.Number of atoms $=

Vedclass · Accepted Answer

$B_1$ has more molecules but equal number of atoms.

Vedclass · Answer

(C) For bulb $B_1$ ($10 \ g$ of $O_2$):Molar mass of $O_2 = 32 \ g/mol$.Number of molecules $= (10 / 32) 	imes N_A = 0.3125 \ N_A$.Number of atoms $= 2 	imes (10 / 32) 	imes N_A = 0.625 \ N_A$.For bulb $B_2$ ($10 \ g$ of $O_3$):Molar mass of $O_3 = 48 \ g/mol$.Number of molecules $= (10 / 48) 	imes N_A = 0.2083 \ N_A$.Number of atoms $= 3 	imes (10 / 48) 	imes N_A = 0.625 \ N_A$.Comparing the results:Number of molecules: $0.3125 \ N_A (B_1) > 0.2083 \ N_A (B_2)$.Number of atoms: $0.625 \ N_A (B_1) = 0.625 \ N_A (B_2)$.Therefore,bulb $B_1$ contains more molecules,and both bulbs contain an equal number of atoms.

Two bulbs $B_1$ and $B_2$ contain $10 \ g$ each of $O_2$ and $O_3$ respectively. Which bulb contains a greater number of molecules and atoms?

Similar Questions

What is the volume occupied by $1$ molecule of water, if its density is $1 \,g \,cm^{-3}$?

The number of $N$ atoms in $681 \ g$ of $C_{7}H_{5}N_{3}O_{6}$ is $x \times 10^{21}$. The value of $x$ is $.....$ $(N_{A} = 6.02 \times 10^{23} \ mol^{-1})$ (Nearest Integer)

Explain: Avogadro's Law.

If $N_A$ is Avogadro’s number,then the number of valence electrons in $4.2 \ g$ of nitride ions $(N^{3 -})$ is ................... $N_A$.

How many protons are present in $1.8 \ g$ of $NH_{4}^{+}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module