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When naphthalene is added to camphor, it disrupts the crystalline structure of camphor by interfering with the intermolecular forces between camphor molecules. This disruption causes the camphor molecules to have a harder time coming together in an organized manner for freezing, resulting in a decrease in the freezing point.
The freezing point is lowered; the decrease is proportional to the content of salt.
Higher the concentration of the solute, lower is the freezing point.
This is a colligative property. Adding a solute will increase the boiling point and decrease the freezing point. The reason has to due with intermolecular forces, and interruption thereof. When water molecules have solute in between them, the temperature has to be lower than normal in order for them to freeze.
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.
The freezing point decrease is -14,8 oC.
To determine the molar mass of a substance using the freezing point depression method, you can measure the decrease in freezing point when a solute is added to a solvent. By knowing the amount of solute added and the decrease in freezing point, you can calculate the molar mass of the solute using the formula: molar mass (mass of solute / moles of solute) (freezing point depression / change in freezing point).
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.
Increasing the concentration of the solute the freezing point decrease.
The freezing point of a solution decreases according to the formula: delta Tf = Kf * molality. Given that the molality of the solution is 4 moles NaCl per kg water and the Kf value for water is 1.86 °C/m, the decrease in freezing point would be approximately 7.44°C.
The freezing point depression constant for water is 1.86°C kg/mol. Using the formula ΔT = i * Kf * molality, where i is the van't Hoff factor (for NaCl, i = 2) and molality = moles of solute / kg of solvent, the freezing point of water would decrease by approximately 14.88°C.
When naphthalene is added to camphor, it disrupts the crystalline structure of camphor by interfering with the intermolecular forces between camphor molecules. This disruption causes the camphor molecules to have a harder time coming together in an organized manner for freezing, resulting in a decrease in the freezing point.
The freezing point of water would decrease if 4 mol of NaCl were added because NaCl is a solute that disrupts the water molecules' ability to form solid ice. Each mole of NaCl added to water reduces the freezing point by approximately 1.86 degrees Celsius. So, with 4 mol of NaCl added, the freezing point of water would decrease by about 7.44 degrees Celsius.
Freezing point depression constants are specific values that depend on the solvent being used. They represent how much the freezing point of a solvent will decrease when a solute is added. The higher the constant, the greater the decrease in freezing point. This means that adding a solute to a solvent will lower the freezing point of the solution compared to the pure solvent.
The melting point of ice decreases when salt is added.
The freezing point is lowered; the decrease is proportional to the content of salt.
The freezing point of water decreases by about 1.86 degrees Celsius for each mole of solute (such as sugar) dissolved in 1 kg of water. So, the freezing point would decrease by 1.86 degrees Celsius for every mole of sugar added.