Find out the total number of electrons presen | vedclass

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

Question

(A) $1$. Molar mass of methane $(CH_4)$ = $12 + (4 	imes 1) = 16 \ g/mol$. $2$. Number of moles of $CH_4$ = $\frac{	ext{Given mass}}{	ext{Molar mas

Vedclass · Accepted Answer

$6.022 	imes 10^{23}$

Vedclass · Answer

(A) $1$. Molar mass of methane $(CH_4)$ = $12 + (4 	imes 1) = 16 \ g/mol$. $2$. Number of moles of $CH_4$ = $\frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{1.6 \ g}{16 \ g/mol} = 0.1 \ mol$. $3$. Number of molecules of $CH_4$ = $0.1 	imes 6.022 	imes 10^{23} = 6.022 	imes 10^{22}$ molecules. $4$. Each molecule of $CH_4$ contains $6$ (from $C$) + $4 	imes 1$ (from $H$) = $10$ electrons. $5$. Total number of electrons = $6.022 	imes 10^{22} 	imes 10 = 6.022 	imes 10^{23}$ electrons.

Find out the total number of electrons present in $1.6 \ g$ of methane $(CH_4)$?

Similar Questions

The number of molecules at $NTP$ in $1 \ mL$ of an ideal gas will be

The mass of $1$ mole of electrons is .......

How many moles of $Mg_{3}(PO_{4})_{2}$ contain $0.25$ moles of $O$ atoms?

Volume occupied by one molecule of water (density $= 1 \ g \ cm^{-3}$) is ...... $ \times 10^{-23} \ cm^3$.

The number of oxygen atoms in $4.4 \ g$ of $CO_2$ is approximately:

Vedclass Test Series

Exam Paper Generator

Online Exam Module