123.0℃ 0.93 atm gas H2O 15.0 L –5.0℃ ice 10.5 g

123.0℃ 0.93 atm gas H2O 15.0 L –5.0℃ ice 10.5 g

123.0℃와 0.93 atm의 수증기(기체 H2O) 15.0 L와 –5.0℃의 얼음 덩어리 10.5 g 중, 어느 시료가 더 많은 분자를 포함하고 있겠는가?

Which sample contains more molecules, 15.0 L of steam (gaseous H2O) at 123.0℃ and 0.93 atm pressure or a 10.5 g ice cube at –5.0℃?

PV = nRT

( 참고 https://ywpop.blogspot.com/2024/05/ideal-gas-equation-pv-nrt.html )

n = PV / RT

= [(0.93) (15.0)] / [(0.08206) (273.15 + 123.0)]

= 0.429 mol

---> 수증기(기체 H2O)의 몰수

H2O의 몰질량 = 18 g/mol 이므로,

n = W / M

( 참고 https://ywpop.tistory.com/7738 )

= 10.5 g / (18 g/mol)

= 0.583 mol

---> 얼음(고체 H2O)의 몰수

답: 얼음. ice

[키워드] Which sample contains more molecules, 15.0 L of steam (gaseous H2O) at 123.0 °C and 0.93 atm pressure or a 10.5 g ice cube at −5.0 °C?