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What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?
The molarity is 2 mol/L.
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 freezing point of water be if 3 mol of sugar were added to 1 kg of water?
-5.58 C
How much would the boiing point of water increase if 4 ml of NaCl were added to 1 kg of water?
(kb=0.51degrees celcious/(mol/kg) for water and i=2 for NaCl)
What is the change in the freezing point of water when 35g of sucrose is dissolved in 300g of water?
To determine the change in the freezing point of water when 35g of sucrose is dissolved in 300g of water, we can use the freezing point depression formula: ΔTf = i * Kf * m, where i is the van 't Hoff factor (1 for sucrose), Kf is the freezing point depression constant for water (1.86 °C kg/mol), and m is the molality of the solution. First, calculate the number of moles of sucrose: ( \text{moles} = \frac{35g}{342.3 g/mol} \approx 0.102 moles ). The mass of the solvent (water) in kg is 0.3 kg, so the molality ( m = \frac{0.102 moles}{0.3 kg} \approx 0.34 , mol/kg ). Thus, the change in freezing point is ( ΔTf = 1 * 1.86 °C kg/mol * 0.34 , mol/kg \approx 0.63 °C ). Therefore, the freezing point of the solution will decrease by approximately 0.63 °C.
What is the molarity of a solution that has mol of kcl in 0.800 kg in water?
4 mol/0.800 kg
What is the molality of a solution that has 6 mol of CaCl2 in 3 kg of water?
2 m
How much would the freezing point of water decrease if 4 mol of sugar were added to 1 kg of water?
7.44 C
What is the molality solution that has 3 mol of glucose in 6 kg if water?
The molality of the solution is calculated by dividing the moles of solute by the mass of the solvent in kg. In this case, there are 3 mol of glucose in 6 kg of water. Therefore, the molality of the solution is 0.5 mol/kg.
What is the molality of a solution that has 3 mol of glucose in 6 kg of water?
The molality of the solution is 0.5 mol/kg. This is calculated by dividing the number of moles of solute (3 mol glucose) by the mass of solvent in kilograms (6 kg water).
What is the molality of a solution made by dissolving 2 mole of NaOH in 6kg of water?
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kg. In this case, there are 2 moles of NaOH dissolved in 6 kg of water. Therefore, the molality would be 2 mol / 6 kg = 0.33 mol/kg.
What is the molarity of a solution that has 6 mol of CaCl2 in 3 kg of water?
The molarity is 2 mol/L.
What is the molality of a solution that has 6 mol of cacl2 in 3kg of water?
The molality of a solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms. In this case, the molality of the CaCl2 solution would be 2 mol/kg, as 6 mol of CaCl2 dissolved in 3 kg of water results in a molality of 2 mol/kg.
What is the molality of solution that has 3 mol of glucose in 6 Kg of water?
3mol/6kg
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 is the molality of .630g of NaCl dissolved in 525g of water?
To find molality, first convert grams of NaCl to moles: 0.630g NaCl / 58.44 g/mol NaCl = 0.0108 mol NaCl. Next, calculate molality using moles of solute and mass of solvent: molality = moles of solute / kg of solvent = 0.0108 mol NaCl / 0.525 kg water = 0.0206 mol/kg.
How much would the freezing point of water decrease if 4 mole of Na Cl were added to 1 kg of water?
The freezing point depression constant for water is 1.86°C kg/mol. With 4 moles of NaCl added to 1 kg of water, the molality would be 4 mol / 1 kg = 4 m. The freezing point depression can be calculated as 1.86°C kg/mol * 4 m = 7.44°C.
