What is the number of moles and total number | vedclass

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(D) Volume of $NH_{3}$ gas at $STP = 5.6 \ cm^{3} = 5.6 	imes 10^{-3} \ L$ (or $dm^{3}$). Since $22.4 \ L$ of any gas at $STP$ contains $1 \ mole$. N

Vedclass · Accepted Answer

$2.50 	imes 10^{-4} \ mol$ and $6.022 	imes 10^{20}$ atoms

Vedclass · Answer

(D) Volume of $NH_{3}$ gas at $STP = 5.6 \ cm^{3} = 5.6 	imes 10^{-3} \ L$ (or $dm^{3}$). Since $22.4 \ L$ of any gas at $STP$ contains $1 \ mole$. Number of moles of $NH_{3} = \frac{5.6 	imes 10^{-3} \ L}{22.4 \ L/mol} = 2.5 	imes 10^{-4} \ mol$. One molecule of $NH_{3}$ contains $4$ atoms ($1$ Nitrogen and $3$ Hydrogen). Total number of atoms = (Number of moles) $	imes$ (Avogadro's number) $	imes$ (Atomicity). Total number of atoms = $2.5 	imes 10^{-4} 	imes 6.022 	imes 10^{23} 	imes 4 = 6.022 	imes 10^{20}$ atoms.

What is the number of moles and total number of atoms respectively present in $5.6 \ cm^{3}$ of ammonia gas at $STP$?

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