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Calculate the heat of sublimation for water when 10 g of ice sublimes at standard conditions (1 atm, 25°C)? Given: Heat of fusion of ice is 6.01 kJ/mol; Heat of vaporization of water is 40.67 kJ/mol.

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To calculate the heat of sublimation for water, we first need to determine the number of moles of water in the 10 g of ice. The molar mass of water is 18.015 g/mol.Number of moles = mass / molar massNumber of moles = 10 g / 18.015 g/mol = 0.555 molesThe heat of sublimation is the sum of the heat of fusion and the heat of vaporization.Heat of sublimation = Heat of fusion + Heat of vaporizationHeat of sublimation = 6.01 kJ/mol + 40.67 kJ/mol = 46.68 kJ/molNow, we can calculate the total heat of sublimation for the 10 g of ice.Total heat of sublimation = Heat of sublimation  Number of molesTotal heat of sublimation = 46.68 kJ/mol  0.555 moles = 25.91 kJTherefore, the heat of sublimation for water when 10 g of ice sublimes at standard conditions is 25.91 kJ.

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