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4 November, 05:33

A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will be vapor when equilibrium is reached? (the vapor pressure of water at 65°C is 187.5 mmHg)

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  1. 4 November, 05:38
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    50.4 % of the water will be vapor

    Explanation:

    Step 1: Data given

    Mass of water = 1.68 grams

    volume of the flask = 5.3 L

    Temperature = 65°C

    Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

    Step 2: Calculate moles of H2O

    p*V=n*R*T

    ⇒ p = the pressure of water = 0.2467 atm

    ⇒ V = the volume of the flask = 5.3 L

    ⇒ n = moles of water

    ⇒ R = gas constant = 0.08206 L*atm / K*mol

    ⇒ T = the temperature = 65°C = 338 Kelvin

    n = (p*V) / (R*T)

    n = (0.2467 * 5.3) / (0.08206 * 338)

    n = 0.047 moles

    Step 3: Calculate mass of water

    Mass of water = moles of water * molar mass of water

    Mass of water = 0.047 moles * 18.02 g/mol

    Mass of water = 0.84694 grams

    Step 4: Calculate the percent of water vaporized

    % = (0.84694 grams/1.68 grams) * 100%

    % = 50.4%

    50.4 % of the water will be vapor
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