The mass of 1 mole electron is 0.55mg
(B) False

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Hint: We know that the mass of an electron is ${\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{ - {\rm{31}}}}\,{\rm{kg}}$. 1 mole of electron is equal to the Avogadro constant that is$6.022 \times {10^{23}}$electrons. Avogadro’s constant is written as ${{\rm{N}}_{\rm{A}}}$

Complete step by step answer:
We know that,
The mass of $1$ electron ${\rm{ = }}\,{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{ - {\rm{31}}}}\,{\rm{kg}}$which is equivalent to ${\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{ - {\rm{25}}}}\,{\rm{mg}}$
1mole of electron = ${\rm{6}}{\rm{.02 \times 1}}{{\rm{0}}^{{\rm{23}}}}{\rm{electrons}}$
For getting the mass of ${\rm{1}}$ mole of electrons we must multiply the mass of an electron with ${\rm{1}}$ mole of electrons
i.e. Mass of 1 mole of electrons ${\rm{ = }}$ mass of $1$ electron $ \times $number of one mole of electron
${\rm{ = }}$$\left( {{\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{ - {\rm{25}}}}} \right){\rm{ \times }}\left( {{\rm{6}}{\rm{.02 \times 1}}{{\rm{0}}^{{\rm{23}}}}} \right)$
${\rm{ = }}$${\rm{0}}{\rm{.54782mg}}$
$ \approx {\rm{0}}{\rm{.55mg}}$
$\therefore $The mass of one mole of electrons ${\rm{ = }}\,{\rm{0}}{\rm{.55mg}}$
Therefore, the mass of $1$ mole of electrons is ${\rm{0}}{\rm{.55mg}}$ this statement is true.

Additional Information:
Mole is a unit to represent the quantity of substance.$1$ mole of any substance contains ${\rm{6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}$ particles and this number is known as Avogadro Number $\left( {{{\rm{N}}_{\rm{A}}}} \right)$.
${\rm{1}}\,{\rm{mol}}\,{\rm{ = }}\,{\rm{6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}$ particles
For finding the number of moles we use the above formula, that is, given mass divided by molar mass. If there is a given number of particles, then we will use the formula given number of particles divided by Avogadro number.
Gram molecular volume: Same volume is occupied by one-gram molecule of any of the dry gas at standard temperature and pressure i.e. ${\rm{22}}{\rm{.4 L}}$ This is termed as gram molecular volume.
Relationship between mole, GMV and Avogadro number can be given as,
1 gram of molecule of any dry gas occupies ${\rm{22}}{\rm{.4 L}}$ volume at STP. Hence at STP${\rm{22}}{\rm{.4 L}}$ of dry gas is equivalent to one mole of that gas and it contains ${\rm{6}}{\rm{.022 \times 1}}{{\rm{0}}^{{\rm{23}}}}$molecules i.e. Avogadro number of molecules.

A mole is the chemist's counting unit. We know that carbon 12 has a significant role in terms of the Avogadro’s constants. The Avogadro constant $6.022 \times {10^{23}}$ is defined as the number of atoms in exactly ${\rm{12g}}$ of carbon ${\rm{12}}$.