The number of moles of electrons in 1 kg of | vedclass

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

Question

(D) The mass of one electron is $m_e = 9.108 	imes 10^{-31} \ kg$.The number of electrons in $1 \ kg$ is $N = \frac{1 \ kg}{9.108 	imes 10^{-31} \ k

Vedclass · Accepted Answer

$\frac{1}{9.108 	imes 6.022} 	imes 10^8$

Vedclass · Answer

(D) The mass of one electron is $m_e = 9.108 	imes 10^{-31} \ kg$.The number of electrons in $1 \ kg$ is $N = \frac{1 \ kg}{9.108 	imes 10^{-31} \ kg} = \frac{1}{9.108} 	imes 10^{31}$.The number of moles of electrons is given by $n = \frac{N}{N_A}$,where $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Therefore,$n = \frac{1}{9.108 	imes 10^{-31}} 	imes \frac{1}{6.022 	imes 10^{23}} = \frac{1}{9.108 	imes 6.022} 	imes 10^{31-23} = \frac{1}{9.108 	imes 6.022} 	imes 10^8$.

The number of moles of electrons in $1 \ kg$ of electrons is .......

Similar Questions

The number of moles of $Mg_3(PO_4)_2$ containing $0.25$ moles of oxygen atoms is:

Number of atoms in $24 \ g$ of $He$ is ............ $N_A$

What is the number of moles of $N$ atoms and number of moles of $O$ atoms respectively present in one mole of uracil?

Which has the highest number of atoms?

$A$ sample of $[Cu(NH_3)_4]SO_4$ contains $2.4 \times 10^{24}$ ammonia molecules. The moles of $[Cu(NH_3)_4]SO_4$ in the given sample will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module