Which one of the following will have the larg | vedclass

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

Question

(C) $(i).$ $1 \,g$ of $Au_{(s)} = \frac{1}{197} \,mol$ of $Au_{(s)} = \frac{6.022 	imes 10^{23}}{197} \approx 3.06 	imes 10^{21}$ atoms of $Au_{(s)}

Vedclass · Accepted Answer

$(iii)$

Vedclass · Answer

(C) $(i).$ $1 \,g$ of $Au_{(s)} = \frac{1}{197} \,mol$ of $Au_{(s)} = \frac{6.022 	imes 10^{23}}{197} \approx 3.06 	imes 10^{21}$ atoms of $Au_{(s)}$.$(ii).$ $1 \,g$ of $Na_{(s)} = \frac{1}{23} \,mol$ of $Na_{(s)} = \frac{6.022 	imes 10^{23}}{23} \approx 26.2 	imes 10^{21}$ atoms of $Na_{(s)}$.$(iii).$ $1 \,g$ of $Li_{(s)} = \frac{1}{7} \,mol$ of $Li_{(s)} = \frac{6.022 	imes 10^{23}}{7} \approx 86.0 	imes 10^{21}$ atoms of $Li_{(s)}$.$(iv).$ $1 \,g$ of $Cl_{2(g)} = \frac{1}{71} \,mol$ of $Cl_{2(g)}$. Since $1 \,mol$ of $Cl_2$ contains $2 \,mol$ of $Cl$ atoms,number of atoms $= \frac{2 	imes 6.022 	imes 10^{23}}{71} \approx 16.96 	imes 10^{21}$ atoms of $Cl$.Comparing the values,$1 \,g$ of $Li_{(s)}$ has the largest number of atoms.

Which one of the following will have the largest number of atoms?
$(i)$ $1 \,g$ $Au_{(s)}$
$(ii)$ $1 \,g$ $Na_{(s)}$
$(iii)$ $1 \,g$ $Li_{(s)}$
$(iv)$ $1 \,g$ of $Cl_{2(g)}$

Similar Questions

Containers $A, B$,and $C$ of equal volume contain oxygen,neon,and methane gases respectively at the same temperature and pressure. Which of the following is the correct order of their increasing mass?

What is the number of oxygen atoms in $8 \ g$ of ozone $(O_3)$?

What is a mole?

Find out the total number of electrons present in $3.2 \ g$ of methane $(CH_4)$?

What is the number of molecules in $4.25 \ g$ of $NH_3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Which one of the following will have the largest number of atoms?$(i)$ $1 \,g$ $Au_{(s)}$$(ii)$ $1 \,g$ $Na_{(s)}$$(iii)$ $1 \,g$ $Li_{(s)}$$(iv)$ $1 \,g$ of $Cl_{2(g)}$