35.0 g C3H7OH 150. g C2H5OH mass percent

35.0 g C3H7OH 150. g C2H5OH mass percent

An alcohol solution contains

35.0 g of 1-propanol (C3H7OH)

and 150. g of ethanol (C2H5OH).

Calculate the mass percent and

the mole fraction of each substance.

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C3H7OH의 %농도

= 35.0 / (35.0 + 150.) × 100

= 18.9%

( 참고 https://ywpop.tistory.com/2656 )

C2H5OH의 %농도

= 150. / (35.0 + 150.) × 100

= 81.1%

C3H7OH의 몰질량

= 3(12) + 8(1) + (16) = 60 g/mol

( 참고 https://ywpop.tistory.com/16840 )

35.0 g / (60 g/mol) = 0.583 mol C3H7OH

( 참고 https://ywpop.tistory.com/7738 )

C2H5OH의 몰질량

= 2(12) + 6(1) + (16) = 46 g/mol

150. g / (46 g/mol) = 3.261 mol C2H5OH

C3H7OH의 몰분율

= 0.583 / (0.583 + 3.261)

= 0.152

( 참고 https://ywpop.tistory.com/2659 )

C2H5OH의 몰분율

= 3.261 / (0.583 + 3.261)

= 0.848

[키워드] C3H7OH 몰분율 기준, C2H5OH 몰분율 기준