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일반화학/[13장] 용액의 성질

137.2 mg protein 100.0 mL water 4°C 6.45 cm

by 영원파란 2022. 10. 1.
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137.2 mg protein 100.0 mL water 4°C 6.45 cm

 

 

단백질 137.2 mg을 4℃ 물 100.0 mL에 녹였더니

물기둥이 6.45 cm 높아졌다.

단백질의 몰질량을 예측하시오.

 

 

When 137.2 mg of protein was dissolved

in 100.0 mL of water at 4°C,

the water column increased by 6.45 cm.

Predict the molar mass of the protein.

 

---------------------------------------------------

 

6.45 cmH2O × (1 m / 100 cm) = 0.0645 mH2O

 

 

 

1 atm = 10.332559 mH2O 이므로,

0.0645 mH2O × (1 atm / 10.332559 mH2O)

= 0.0062424 atm

 

 

 

π = MRT

( 참고 https://ywpop.tistory.com/1921 )

 

M = π / RT

= (0.0062424) / [(0.08206) (273.15 + 4)]

= 0.0002744765 mol/L

 

 

 

(0.0002744765 mol/L) (100.0/1000 L)

= 2.744765E–5 mol

 

 

 

(137.2/1000 g) / (2.744765E–5 mol)

( 참고 https://ywpop.tistory.com/7738 )

 

= 4998.61 g/mol

 

 

 

답: 약 5000 g/mol

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/22368 ] 100 mg protein water 10.0 mL osmotic 13.3 mmHg 25℃

 

 

 

[키워드] 삼투압 몰질량 기준문서

 

 

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