2.05 g polystyrene toluene 0.100 L osmotic 25℃ 1.21 kPa

A sample of 2.05 g of polystyrene with a uniform polymer chain length was dissolved in enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25℃. Calculate the molar mass of the polystyrene.

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

1 atm = 101.325 kPa 이므로,

1.21 kPa × (1 atm / 101.325 kPa)

= 0.011942 atm

π = MRT

( 참고 https://ywpop.tistory.com/1921 )

M = π / RT

= (0.011942) / (0.08206 × (273.15 + 25))

= 0.0004881 mol/L

---> 용액의 몰농도

용액 속 polystyrene의 몰수

= (0.0004881 mol/L) (0.100 L)

= 4.881×10^(-5) mol

( 참고 https://ywpop.tistory.com/7787 )

polystyrene의 몰질량, M

= W / n

( 참고 https://ywpop.tistory.com/7738 )

= 2.05 g / (4.881×10^(-5) mol)

= 41999.59 g/mol

= 42000 g/mol

답: 약 4.20×10^4 g/mol

[키워드] 폴리스티렌 삼투압 기준, 폴리스티렌 몰질량 기준, 폴리스타이렌 삼투압 기준, 폴리스타이렌 몰질량 기준

2.05 g polystyrene toluene 0.100 L osmotic 25℃ 1.21 kPa