The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
The change in freezing point is -0.430°C. The molal freezing point depression constant for water is 1.86°C kg/mol. Calculate the molality of the solution using the formula: ΔTf = Kf * m. 8.02 g of solute in 861 g of water is equivalent to 8.02 / 58.44 = 0.1373 mol of solute.
Freezing point.
Lithium fluoride has the molecular formula of LiF. LiF has a molecular weight of 25.94 grams per mole and melting point of 845 degrees Celsius.
The freezing cannot be stopped; only the freezing temperature is lowered adding salts.
It takes more heat to vaporize 12 grams of CH4 (methane) compared to 12 grams of Hg (mercury) because methane has weaker intermolecular forces and a lower boiling point. This means more energy is required to break the bonds between methane molecules to allow them to vaporize. Mercury has stronger intermolecular forces, so it requires less energy to vaporize.
The freezing point of solution is always less than that of the freezing point of the pure solvent. The freezing point of pure water is 0 (zero) degree celsius. The freezing point of the water decreases with the increase in the sugar concentration. for ex. a 10 grams of sugar when dissolved in 100 grams of water, the freezing point depression of -0.56 degree Celsius A 10 molal sucrose will bring about the depression in freezing point of water to about -20 degree celsius
To decrease the freezing point of water by 2.5 degrees Celsius, you would need to dissolve approximately 37.5 grams of sugar in 300 grams of water. This is calculated based on the colligative property that states freezing point depression is directly proportional to the molality of the solute in the solution.
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
The freezing point of sodium permanganate is 36 degrees Celsius. This is an inorganic compound that has the chemical formula of NaMnO4. Its molar mass is 141.9254 grams per mole.
To calculate the molecular mass of the solute, we need to use the formula for freezing point depression: ΔTf = Kf * m. Given that the ΔTf is -0.430°C, the molal concentration (m) of the solute can be found by dividing the grams of solute by the grams of water. Substituting these values into the formula allows us to solve for Kf, which can eventually be used to determine the molecular mass of the solute.
A refrigerant with a low boiling point allows it to easily absorb heat from the surroundings and vaporize, transferring heat effectively. A low freezing point ensures that the refrigerant remains in a liquid state in sub-zero temperatures, preventing damage to the refrigeration system.
The change in freezing point is -0.430°C. The molal freezing point depression constant for water is 1.86°C kg/mol. Calculate the molality of the solution using the formula: ΔTf = Kf * m. 8.02 g of solute in 861 g of water is equivalent to 8.02 / 58.44 = 0.1373 mol of solute.
This quantity is equivalent to 90 g glucose / kg water = 0.50 mole particles of solute / kg water, so with a 'molar cryoscopic constant' for water of -1.86 oC/kgthis lowers the freezing point to -0.93 oC.
Freezing point.
5.04
The freezing cannot be stopped; only the freezing temperature is lowered adding salts.