For questions 2 to 4 d, consider two 2.00 L flasks, one of which contains 0.250

Question

For questions 2 to 4 d, consider two 2.00 L flasks, one of which contains 0.250 mole of O2

gas at 50.0 deg C and the other of which contains 0.500 mole of H2 gas at 50.0 deg C.

2.      Which sample contains more molecules? Show your calculation. (6 pt)

3.      Which sample has the greater mass? Show your calculation. (6 pt)

4.      a.      In which flask is the ratio of molecular volume (volume occupied by the

                  molecules touching one another in a condensed state) to gaseous sample

                  volume (volume occupied by the entire gaseous sample) greater? (2 pt)

         b.      What is the approximate value of these ratios? Explain. Calculation

                  not needed (4 pt)

         c.      In which sample do the molecules have higher average kinetic energy? Explain.

                  Calculation not needed (4 pt)

         d.      In which flask do the molecules have higher average speed? Explain.

                  Calculation not needed (4 pt)

Explanation / Answer

Two flasks :

First have 0.250 mol O2

Second has 0.500 mol H2

2. Number of molecules of O2 in flask = 0.250 mol x 6.023 x 10^23 = 1.50 x 10^23 molecules

Number of molecules of H2 in flask = 0.500 mol x 6.023 x 10^23 = 3.01 x 10^23 molecules

So, the number of molecules of H2 are greater than O2 in the flask.

3. O2 has higher molar mass so, mass of flask with O2 is heavier than the one with H2 in it.

4. a. Higher moles means greater molecular volume in this case. Here, H2 would be more in gaseous state and hence the ratio of volumes in condensed to gaseous state would be lower. Thus molecular volume for O2 would be higher.

b. Approximate ratios of the molecular volumes of O2/H2 = 2/1

c. Higher velocity molecule would have higher average kinetic energy, so smaller molecule H2 would have greater average kinetic energy as compared to larger and slower molecule O2 in the flask.

d. average speed relation (v),

v = sq.rt.(3RT/molar mass (kg))

with 3RT term constant in both O2 and H2

we can calculate approximate speeds

So,

v(O2) = sq.rt.(1/32 x 10^-3) = 5.59

v(H2) = sq.rt.(1/2 x 10^-3) = 22.36

So, as can be seen, average speed of H2 will be much greater than O2.

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