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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 molecular mass of the nonionic solute called 8.02 grams of solute in 861 grams of water lower the freezing point to -0.430 C?
To find the molecular mass of the nonionic solute, we can use the freezing point depression formula: ΔTf = Kf * m, where ΔTf is the change in freezing point, Kf is the freezing point depression constant for water (1.86 °C kg/mol), and m is the molality. The change in freezing point is 0.430 °C, so we can rearrange the formula to find molality: m = ΔTf / Kf = 0.430 °C / 1.86 °C kg/mol ≈ 0.231 mol/kg. Since molality (m) is moles of solute per kg of solvent, and we have 0.231 moles in 0.861 kg of water, we can calculate the number of moles: 0.231 mol/kg * 0.861 kg ≈ 0.199 moles. Finally, the molecular mass is given by the mass of the solute divided by the number of moles: 8.02 g / 0.199 mol ≈ 40.3 g/mol.
How does salt affect the melting and freezing point of water?
Salt decreases the freezing point of water and increases the boiling point of water.
In a titration when the number of moles of hydrogen ions equals the number of moles of hydroxide ions what is said to have happened?
When the number of moles of hydrogen ions equals the number of moles of hydroxide ions in a titration, it means that the solution has reached the equivalence point. At this point, the solution is neutral and the pH is typically around 7, indicating that the acid and base have reacted completely with each other.
How does the presence of dissolved sodium chloride affect the boiling and freezing point of water?
Dissolved solute (NaCl, salt) will raise the boiling point and lower the freezing point of water. This is known as a colligative property.
What general statement can you make concerning the relationship between the number of moles of solute and the magnitude of the freezing point depression?
Increasing the concentration of the solute the freezing point decrease.
Would vinegar affect the freezing point?
Vinegar will not affect the freezing point of vinegar.
Which would lower the freezing point more 100 moles of sugar or 100 moles of NaCl?
100 moles of NaCl would lower the freezing point more than 100 moles of sugar. This is because NaCl dissociates into more particles in solution compared to sugar, resulting in a greater depression of the freezing point due to colligative properties.
What is the freezing point of potassium hydroxide?
48% KOH freezing pt -11deg C 45% KOH freezing point -28 deg C The change in freezing point (always a decrease) = (number of ions in solution per molecule) x (Kf - the freezing point constant of the solvent) x (m - the molality of the solution, i. e. moles solute per kg solvent) For KOH in water, Freezing pt = 0 - 2(1.86)(molality of solution)
What is the freezing point of a solution that contains 0.550 moles of Nal in 615 g of water?
The freezing point depression equation is used to calculate the freezing point of a solution. Given the molality of the NaI solution and the molecular weight of water, the freezing point of the solution can be determined.
What is the freezing point of a solution that contains 0.550 moles of NaI in 615 g of water?
To calculate the freezing point depression, you first need to find the molality of the solution using the moles of solute and mass of solvent. Then, use the molality to find the freezing point depression constant of water. Finally, apply the formula ΔTf = Kf * molality to find the freezing point depression.
Freezing point is the point at which a liquid turns into a solid What effect can change the freezing point?
Changing the pressure can affect the freezing point of a substance. Generally, an increase in pressure will lower the freezing point, while a decrease in pressure will raise the freezing point. The presence of solutes or impurities in the liquid can also change the freezing point.
How much would the freezing point of water decrease if 4 moles of Na-Cl were added to 1 Kg of water?
The freezing point decrease is -14,8 oC.
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 molecular mass of the nonionic solute called 8.02 grams of solute in 861 grams of water lower the freezing point to -0.430 C?
To find the molecular mass of the nonionic solute, we can use the freezing point depression formula: ΔTf = Kf * m, where ΔTf is the change in freezing point, Kf is the freezing point depression constant for water (1.86 °C kg/mol), and m is the molality. The change in freezing point is 0.430 °C, so we can rearrange the formula to find molality: m = ΔTf / Kf = 0.430 °C / 1.86 °C kg/mol ≈ 0.231 mol/kg. Since molality (m) is moles of solute per kg of solvent, and we have 0.231 moles in 0.861 kg of water, we can calculate the number of moles: 0.231 mol/kg * 0.861 kg ≈ 0.199 moles. Finally, the molecular mass is given by the mass of the solute divided by the number of moles: 8.02 g / 0.199 mol ≈ 40.3 g/mol.
How does salt affect the melting and freezing point of water?
Salt decreases the freezing point of water and increases the boiling point of water.
Does mass affect freezing point?
Well, I did an experiment in class on this question. We used different amounts lauric acid and it turned out that the freezing point was pretty much the same for all the samples. So, in all, the freezing point does not depend on the mass of a substance.
