8. A student collected the following data: Temperature of the water bath 90.0°C,

Question

8. A student collected the following data: Temperature of the water bath 90.0°C, .OUJ Of barometric pressure = 765.0 torr, and the volume of the flask-260.0 mL Un- known to the student, during the experiment some unvapori mained in the flask. If his unknown gas was dichloroethane (C,H,Cl2), and his calculated molar mass was 112 g/mol, what mass of liquid remained un- vaporized? zed liquid re- 9. A student collected the following data: Mass of gas 0.800 g, temperature of the water bath 95.0°C, barometric pressure 760.0 torr, and the volume of the flask = 250.0 ml . Later the student was told that one drop (0.050g) of water had been present in the flask during the experiment and it completely vaporized contaminating the gas. (a) Calculate the apparent molar mass, that is, the molar mass the student would get if he did not know that there were any impurities present. (b) Show how the student could still calculate the true molar mass of the gas. (c) Calculate the percentage of error in part (a).

Explanation / Answer

Answer:

8)

PV = nRT

or, PV = (g/M) RT

Now had the whole amount vaporize, calculated molar mass would be 99g/mol. Use this value to calculate total mass of vapor present.

PV = gRT/M

or, g = PV * M/RT

        = 1 atm * 0.26 L*99g/mol /0.082 L atm /K/mol * 363 K = 0.865 g

But the student has calculated the molar mass to be 112g/mol. Mass of the vapor .

g = PVM/RT

    = 1 atm * 0.26 L*112g/mol /0.082 L atm /K/mol * 363 K = 0.978 g

Mass of liquid remained unvaporized = 0.978g-0.865 g= 0.1133 g

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