1) A This is part d of analysis in procedure 1 on page 5. The density of uncompr

Question

1) A

This is part d of analysis in procedure 1 on page 5. The density of uncompressed air is 1.293 kg/m3, and one mole of air (containing Avogadro's number of molecules) has a mass of 28.94 g.

1-a) Use the density and the volume of the uncompressed air 45 cc to determine the mass of the air in the syringe, mair, in grams. (You will have to do some unit conversions.)

mair = ______ g

1) B

Now use the mass of one mole of air and Avogadro's number to determine N, the number of molecules in the syringe. ______________________

1) C

In procedure 1, suppose the volume of the syringe is 30.9 cc. You measure the pressure as 1.55 atm, and the temperature as 23.4 Celsius. Assume: the diameter of the plunger on the syringe is 1.94 cm.

Find
- N, the number of molecules of gas in the syringe: __________ molecules

Here is the extra info if needed:

d) Now, we will see how well the experiment worked. The density of uncompressed air is 1.293 kg/m3, and one mole of air (containing Avogadro’s number of molecules) has a mass of 28.94 g.   In the space provided: • The air was uncompressed when the volume of your syringe was 45 cc. Use the density and the volume of the uncompressed air to determine the mass of the air in the syringe, in grams. (You will, of course, have to do some unit conversions.) • Use the mass of one mole of air and Avogadro’s number to determine N, the number of molecules in the syringe. This is the calculated value for N.

Explanation / Answer

a)

mass = density*Volume

mass = 1.293*1000*4.5*10^-5

mass = 0.058 grams

b)

Number of molecules are 6.023*10^23 * 0.058/28.94

= 12*10^20 MOLECULES

C)

pv = nRT

n = pv/RT = 1.55 *101325*30.9*10^-6/8.314*296.4

n = 0.00196

Number of molecules are

N = 0.00196*6.023*10^23

N = 11.86*10^20 Molecules

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