If 500 mL of NH_3 gas at STP contains 6.0 ti | vedclass

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(C) According to Avogadro's Law,at constant temperature and pressure,the number of molecules is directly proportional to the volume of the gas $(N \pr

Vedclass · Accepted Answer

$1.2 	imes 10^{21}$

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(C) According to Avogadro's Law,at constant temperature and pressure,the number of molecules is directly proportional to the volume of the gas $(N \propto V)$.Therefore,$\frac{N_1}{N_2} = \frac{V_1}{V_2}$.Given: $V_1 = 500 \ mL$,$N_1 = 6.0 	imes 10^{21}$ molecules,$V_2 = 100 \ mL$.$\frac{6.0 	imes 10^{21}}{N_2} = \frac{500 \ mL}{100 \ mL}$.$N_2 = \frac{6.0 	imes 10^{21} 	imes 100}{500} = \frac{6.0 	imes 10^{21}}{5} = 1.2 	imes 10^{21}$ molecules.

If $500 \ mL$ of $NH_3$ gas at $STP$ contains $6.0 \times 10^{21}$ molecules,how many molecules will $100 \ mL$ of $CO_2$ gas contain at $STP$?

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