1/5 1 At 24.55 °C the vapor pressure of pure water is PH20 =54.07 mmHg. Calculate the amount (in grams) of a non-dissociating solute (molar mass = 137.18) that when added to 286.852 g of water will reduce the vapor pressure by 5.45 mmHg when compared with pure water at the same temperature. (The answer should be given with three numbers after the decimal; example 25.143) Molar mass of water = 18.02 g/mol R= 0.08206 atm-L/mol-K 0 °C = 273.15 K 1.0000 atm = 760.00 mmHg

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At 24.55 °C the vapor pressure of pure water is PH20 = 54.07 mmHg. Calculate the amount (in grams) of a non-dissociating solute (molar mass = 137.18) that when added to
286.852 g of water will reduce the vapor pressure by 5.45 mmHg when compared with pure water at the same temperature.
(The answer should be given with three numbers after the decimal; example 25.143)
Molar mass of water = 18.02 g/mol
R = 0.08206 atm-L/mol-K
0 °C = 273.15 K
1.0000 atm =760.00 mmHg
Type your answer..
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Transcribed Image Text:1/5 At 24.55 °C the vapor pressure of pure water is PH20 = 54.07 mmHg. Calculate the amount (in grams) of a non-dissociating solute (molar mass = 137.18) that when added to 286.852 g of water will reduce the vapor pressure by 5.45 mmHg when compared with pure water at the same temperature. (The answer should be given with three numbers after the decimal; example 25.143) Molar mass of water = 18.02 g/mol R = 0.08206 atm-L/mol-K 0 °C = 273.15 K 1.0000 atm =760.00 mmHg Type your answer.. Next
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