본문 바로가기
반응형

일반화학/[13장] 용액의 성질141

15.4 g 요소 + 66.7 mL water 용액의 끓는점과 어는점 15.4 g 요소 + 66.7 mL water 용액의 끓는점과 어는점 What are the normal freezing points and boiling points of the following solutions? (b) 15.4 g urea in 66.7 mL water --------------------------------------------------- ▶ 참고: 끓는점 오름과 어는점 내림 [ https://ywpop.tistory.com/1920 ] --------------------------------------------------- urea, CO(NH2)2의 몰질량 = 60.06 g/mol 이므로, 15.4 g / (60.06 g/mol) = 0.25641 mol CO(NH2).. 2022. 10. 12.
21.2 g NaCl + 135 mL water 용액의 끓는점과 어는점 21.2 g NaCl + 135 mL water 용액의 끓는점과 어는점 What are the normal freezing points and boiling points of the following solutions? (a) 21.2 g NaCl in 135 mL water --------------------------------------------------- ▶ 참고: 끓는점 오름과 어는점 내림 [ https://ywpop.tistory.com/1920 ] --------------------------------------------------- NaCl의 몰질량 = 58.44 g/mol 이므로, 21.2 g / (58.44 g/mol) = 0.362765 mol NaCl (용질) ( 참고 h.. 2022. 10. 12.
왜 Ce2(SO4)3의 용해도는 온도가 증가할 때 감소하는가? 왜 Ce2(SO4)3의 용해도는 온도가 증가할 때 감소하는가? Why does Ce2(SO4)3 decrease in solubility as the temperature increases? --------------------------------------------------- ▶ 참고: 용해도에 영향을 주는 인자 [ https://ywpop.tistory.com/15288 ] --------------------------------------------------- Ce2(SO4)3 becomes less soluble in water as temperature increases because it has a negative entropy of solvation. Ce2(SO4)3는 용매화 엔트.. 2022. 10. 11.
염화마그네슘 0.952 g을 물 100 g에 녹인 용액의 삼투압 염화마그네슘 0.952 g을 물 100 g에 녹인 용액의 삼투압 A 0.952 g sample of magnesium chloride dissolves in 100. g of water --------------------------------------------------- MgCl2의 몰질량 = 95.21 g/mol 이므로, 0.952 g / (95.21 g/mol) = 0.0100 mol MgCl2 용액의 질량 = 0.952 g + 100. g = 100.952 g 용액의 밀도 = 물의 밀도 = 1 g/mL 라 가정하면, 용액의 부피 = 100.952 mL ≒ 101 mL 몰농도 = 용질 mol수 / 용액 L수 = 0.0100 mol / (101/1000 L) = 0.0990 mol/L 온도 = .. 2022. 10. 3.
끓는점오름. PABA 8.3 g 62.62℃ 150 g 61.15℃ Kb 3.63 ℃/m 끓는점오름. PABA 8.3 g 62.62℃ 150 g 61.15℃ Kb 3.63 ℃/m 8.32 g PABA 150 g chloroform boil 62.62℃ 61.15℃ Kb 3.63 ℃/m 8.32 g PABA와 150 g CHCl3를 포함하는 용액은 62.62℃에서 끓는다. 같은 압력에서, 순수한 CHCl3는 61.15℃에서 끓는다. PABA의 분자량은 얼마인가? 단, CHCl3의 Kb = 3.63 ℃/m. A solution containing 8.32 g of para-aminobenzoic acid (PABA) and 150 g of chloroform boils at 62.62℃. At the same pressure, pure chloroform boils at 61.15℃. What .. 2022. 10. 3.
137.2 mg protein 100.0 mL water 4°C 6.45 cm 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.. 2022. 10. 1.
100 mg protein water 10.0 mL osmotic 13.3 mmHg 25℃ 100 mg protein water 10.0 mL osmotic 13.3 mmHg 25℃ 100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mmHg at 25℃, what is the molar mass of the protein? --------------------------------------------------- π = MRT ( 참고 https://ywpop.tistory.com/1921 ) M = π / RT = (13.3/760) / [(0.08206) (273.15 + 25)] = 0.00071527 .. 2022. 10. 1.
boiling point 102.36℃ water 250 g 180.9 g mol^-1 glucose gram boiling point 102.36℃ water 250 g 180.9 g mol^-1 glucose gram How many grams of glucose (molar mass = 180.9 g/mol) must be dissolved in 250 g of water to raise the boiling point to 102.36℃? --------------------------------------------------- 끓는점오름, ΔTb ΔTb = Kb × m ( 참고 https://ywpop.tistory.com/1920 ) m = ΔTb / Kb = 2.36 / 0.51 = 4.62745 mol/kg 몰랄농도 = 용질 mol수 / 용매 kg수 이므로, 4.62745 mol/kg = ? mo.. 2022. 10. 1.
128.2 g mol^-1 5.0% w/w 77.3℃ 80.2℃ Kf = 6.9℃ m^-1 128.2 g mol^-1 5.0% w/w 77.3℃ 80.2℃ Kf = 6.9℃ m^-1 A mixture is prepared that is 5.0% (w/w) of an unknown substance mixed with naphthalene (molar mass = 128.2 g mol^-1). The freezing point of this mixture is found to be 77.3℃. What is the molar mass of the unknown substance? The melting point of naphthalene is 80.2℃ and Kf = 6.9℃ m^-1. --------------------------------------------------- 어는점내림, Δ.. 2022. 10. 1.
0.091 m CaCl2 0.440℃ 해리 백분율 0.091 m CaCl2 0.440℃ 해리 백분율 The freezing-point depression of a 0.091 m solution of CaCl2 is 0.440 °C. --------------------------------------------------- 어는점내림, ΔTf ΔTf = i × Kf × m ( i = van't Hoff 인자 ) ( 참고: https://ywpop.tistory.com/1920 ) CaCl2(aq) → Ca^2+(aq) + 2Cl^-(aq) ... i = 3 ---> 100% 이온화된다고 가정하면, i = 3 i = ΔTf / (Kf × m) = 0.440 / (1.86 × 0.091) = 2.60 해리 백분율을 계산하면, (2.6 / 3) × 100 =.. 2022. 10. 1.
반응형