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일반화학/[13장] 용액의 성질141

C12H10 5.50 g benzene 100.0 g 0.903℃ 6.30 g 150.0 g 0.597℃ C12H10 5.50 g benzene 100.0 g 0.903℃ 6.30 g 150.0 g 0.597℃ (a) When 5.50 g of biphenyl (C12H10) is dissolved in 100.0 g of benzene (C6H6), the boiling point increases by 0.903℃. Calculate the Kb for benzene. (b) If 6.30 g of an unknown were added to 150.0 g of benzene, the boiling point of the solution increases by 0.597℃. What is the molar mass of the unknown substance? ------------------------.. 2022. 11. 2.
25℃ benzene vapor 0.1252 atm 6.40 g C10H8 78.0 g benzene 25℃ benzene vapor 0.1252 atm 6.40 g C10H8 78.0 g benzene At 25℃, the vapor pressure of pure benzene is 0.1252 atm. When 6.40 g of a solid sample of naphthalene, C10H8 (128.17 g/mol), is dissolved in 78.0 g of pure benzene, C6H6 (78.0 g/mol), calculate the vapor pressure of benzene above the solution. 25℃에서 순수한 벤젠의 증기압은 0.1252 atm이다. 나프탈렌 (C10H8) 6.40 g을 벤젠 78.0 g에 녹였다. 용액 위의 벤젠의 증기압은 얼마인가? -----.. 2022. 11. 2.
25℃ benzene vapor 0.1252 atm 6.50 g C10H8 76.0 g benzene 25℃ benzene vapor 0.1252 atm 6.50 g C10H8 76.0 g benzene At 25℃, the vapor pressure of pure benzene is 0.1252 atm. When 6.50 g of a solid sample of naphthalene, C10H8 (128.17 g/mol), is dissolved in 76.0 g of pure benzene, C6H6 (78.0 g/mol), calculate the vapor pressure of benzene above the solution. --------------------------------------------------- 각 물질의 몰수를 계산하면, 6.50 g / (128.17 g/mol) = 0... 2022. 11. 2.
fructose 1.049 g/cm3 osmotic 17.0 atm 25℃ freeze fructose 1.049 g/cm3 osmotic 17.0 atm 25℃ freeze An aqueous fructose solution having a density of 1.049 g/cm3 is found to have an osmotic pressure of 17.0 atm at 25℃. Find the temperature at which this solution freezes. Given: for water Kf = 1.86 ℃/m; molecular mass of fructose = 180.16 g/mol A) –1.30℃ B) –1.41℃ C) –1.52℃ D) –1.57℃ E) –1.69℃ --------------------------------------------------- 삼투.. 2022. 10. 22.
sucrose fructose 2.850 g 1.50 L 298.0 K 0.1843 atm osmotic sucrose fructose 2.850 g 1.50 L 298.0 K 0.1843 atm osmotic When a 2.850 g mixture of the sugars sucrose (C12H22O11) and fructose (C6H12O6) was dissolved in water to a volume of 1.50 L, the resultant solution gave an osmotic pressure of 0.1843 atm at 298.0 K. What is X_sucrose of the mixture? --------------------------------------------------- 삼투압, π = MRT ( 참고 https://ywpop.tistory.com/1921 ) M .. 2022. 10. 19.
61.0℃ benzene vapor pressure 54.0 kPa 500 g 51.0 g 49.0 kPa molar mass 61.0℃ benzene vapor pressure 54.0 kPa 500 g 51.0 g 49.0 kPa molar mass 61.0℃에서 벤젠의 증기압은 54.0 kPa이다. 500 g 벤젠에 51.0 g 비휘발성 화합물을 녹였더니 증기압이 49.0 kPa로 떨어졌다. 이 화합물의 몰질량을 구하시오. The vapor pressure of benzene is 54.0 kPa at 61.0℃, but it fall to 49.0 kPa when 51.0 g nonvolatile organic compound is dissolved in 500 g benzene. The molar mass of the nonvolatile compound is ---------------------------------.. 2022. 10. 14.
60.3℃ benzene vapor pressure 53.3 kPa 500 g 19.0 g 51.5 kPa molar mass 60.3℃ benzene vapor pressure 53.3 kPa 500 g 19.0 g 51.5 kPa molar mass 60.3℃에서 벤젠의 증기압은 53.3 kPa이다. 500 g 벤젠에 19.0 g 비휘발성 화합물을 녹였더니 증기압이 51.5 kPa로 떨어졌다. 이 화합물의 몰질량을 구하시오. The vapor pressure of benzene is 53.3 kPa at 60.3℃, but it fall to 51.5 kPa when 19.0 g non-volatile organic compound is dissolved in 500 g benzene. The molar mass of the non-volatile compound is -------------------------------.. 2022. 10. 14.
5.50 g substance 125 g CCl4 boiling point 0.794 K molar mass 5.50 g substance 125 g CCl4 boiling point 0.794 K molar mass 어떤 물질 5.50 g을 사염화탄소 125 g에 녹였더니 끓는점이 0.794 K 상승하였다. 5.50 g of a substance dissolved in 125 g of CCl4 leads to an elevation of the boiling point of 0.794 K. Calculate the freezing point depression, the molar mass of the substance, and the factor by which the vapor pressure of CCl4 is lowered. --------------------------------------------.. 2022. 10. 14.
4.50 g substance 125 g CCl4 boiling point 0.650 K molar mass 4.50 g substance 125 g CCl4 boiling point 0.650 K molar mass 어떤 물질 4.50 g을 사염화탄소 125 g에 녹였더니 끓는점이 0.650 K 상승하였다. 4.50 g of a substance dissolved in 125 g of CCl4 leads to an elevation of the boiling point of 0.650 K. Calculate the freezing point depression, the molar mass of the substance, and the factor by which the vapor pressure of CCl4 is lowered. --------------------------------------------.. 2022. 10. 14.
redwoods 105 m 350 ft water osmotic pressure redwoods 105 m 350 ft water osmotic pressure The tallest trees known are the redwoods in California. Assuming the height of a redwood to be 105 m (about 350 ft), estimate the osmotic pressure required to push water up from the roots to the tree top. --------------------------------------------------- 물의 밀도 = 1 g/mL 라 가정하면, (1 g/mL) (1 kg / 1000 g) (1000 mL / 1 L) (1000 L / 1 m3) = (1) (1 / 1000).. 2022. 10. 12.
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