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Will 1 mol of sugar have the same effect as 1 mol of table salt in lowering the freezing point of water?
no because the mol of sugar is 34 g while salt's mol is 5.8 grams
How much would the freezing point of water decrease if 4 mol of sugar were added to 1 kg of water?
7.44 C
How much would the boiling point of water increase if 4 mol of NaCl were added to 1kg of water?
To determine the boiling point elevation of water when 4 mol of NaCl is added to 1 kg of water, we can use the formula: ΔT_b = i * K_b * m, where ΔT_b is the boiling point elevation, i is the van 't Hoff factor (which is 2 for NaCl), K_b for water is approximately 0.512 °C kg/mol, and m is the molality. Since NaCl dissociates into 2 ions (Na⁺ and Cl⁻), i = 2, resulting in a total of 8 mol of particles (4 mol NaCl × 2). The molality (m) is 4 mol / 1 kg = 4 mol/kg. Thus, ΔT_b = 2 * 0.512 °C kg/mol * 4 mol/kg = 4.096 °C. Therefore, the boiling point of water would increase by approximately 4.1 °C.
How much would the boiling point of water increase if 4 mol of NaCl were addes to 1 kg of water?
The boiling point elevation constant for water is 0.512 °C/kg/mol. When 4 mol of NaCl are added to 1 kg of water, the increase in boiling point would be 4 * 0.512 = 2.048 °C.
What would the final freezing point of water be if 3 mol of sugar were added to kg of water (Kf 1.86?
The freezing point depression can be calculated using the formula ΔTf = Kf * m, where Kf is the freezing point depression constant (1.86 °C/kg) and m is the molality of the solution. With 3 mol of sugar dissolved in 1 kg of water, the molality is 3 mol / 1 kg = 3 mol/kg. Thus, the freezing point depression would be ΔTf = 1.86 °C/kg * 3 mol/kg = 5.58 °C. The final freezing point of the solution would be the freezing point of water (0°C) minus the freezing point depression, so the final freezing point would be -5.58°C.
How much would the boiling point of water increase if 4 mol of NaCl were added to 1 kg of water?
4.08 degrees celcius
How much would the freezing point of water decrease if 4 mol of sugar were added to 1kg of water?
The normal freezing temperature for pure water is 0c. Howeverif sugar is added in the pure water, the freezing point will be lower than zero. How far below zero will depend on the sugar concentration in the water.
How much heat is required to vaporize 43.9 grams of acetone at its boiling point?
molar masse acetone: 58.08 g/mol 43.9g/58.08g/mol =0.75585mol the energy required for vaporization to a gas is... 0.75585molx29.1KJ/mol =21.995KJ
At the equivalence point when exactly 0.050 mol of HCl (aq) has reacted with 0.050 mol of AgNO3 (aq) would you expect the solution to conduct electricity?
Of course - yes.
The freezing point of benzene is 5.5C What is the freezing point of a solution of 5.00 g of naphthalene C10H8 in 444g of benzene Kf of benzene 4.90Cm?
ΔTf = Kf * molality Molar mass of naphthalene = 128.17 g/mol Moles of naphthalene = 5.00 g / 128.17 g/mol = 0.039 mol Molality = moles of naphthalene / kg of solvent = 0.039 mol / 0.444 kg = 0.088 mol/kg ΔTf = 4.90°C/m * 0.088 mol/kg = 0.432°C Freezing point of benzene is 5.5°C, so the freezing point of the solution is 5.5°C - 0.432°C = 5.068°C.
What is the freezin of 0.5 mol of Li Br in 500 ml if water?
To determine the freezing point depression of a solution, we can use the formula: ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van 't Hoff factor (which is 2 for LiBr because it dissociates into Li⁺ and Br⁻), Kf is the freezing point depression constant for water (approximately 1.86 °C kg/mol), and m is the molality of the solution. First, we calculate the molality: 0.5 mol of LiBr in 0.5 kg of water (since 500 ml of water has a mass of about 0.5 kg) gives a molality of 1 mol/kg. Therefore, ΔTf = 2 * 1.86 °C kg/mol * 1 mol/kg = 3.72 °C. The freezing point of pure water is 0 °C, so the freezing point of the solution would be approximately -3.72 °C.
What would the final freezing point of water be if 3mol of NaCl were added to 1kg of water?
The freezing point depression constant for water is 1.86°C kg/mol. First, calculate the molality of the solution: 3 mol NaCl / 1 kg H2O = 3 mol/kg. Next, calculate the freezing point depression: ΔTf = iKfm where i is the van't Hoff factor (2 for NaCl), Kf is the freezing point depression constant, and m is the molality. Plugging in the values, the final freezing point would be -11.16°C.
