Calculate the volume of 6.022 times 10^{21} m | vedclass

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Question

(A) Given: Number of molecules $N = 6.022 	imes 10^{21}$.Number of moles $n = \frac{N}{N_A} = \frac{6.022 	imes 10^{21}}{6.022 	imes 10^{23}} = 0.0

Vedclass · Accepted Answer

$124.7$

Vedclass · Answer

(A) Given: Number of molecules $N = 6.022 	imes 10^{21}$.Number of moles $n = \frac{N}{N_A} = \frac{6.022 	imes 10^{21}}{6.022 	imes 10^{23}} = 0.01 \ mol$.Temperature $T = 300 \ K$,Pressure $P = 2 \ bar$,$R = 8.314 	imes 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1}$.Using ideal gas equation $PV = nRT$:$V = \frac{nRT}{P} = \frac{0.01 	imes 8.314 	imes 10^{-2} 	imes 300}{2}$.$V = \frac{0.24942}{2} = 0.12471 \ L$.Converting to $mL$: $0.12471 	imes 1000 = 124.71 \ mL$.

Calculate the volume of $6.022 \times 10^{21}$ molecules of $CO_2$ at $300 \ K$ temperature and $2 \ bar$ pressure. $[R = 8.314 \times 10^{-2} \ bar \ L \ K^{-1} \ mol^{-1}]$ (in $mL$)

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