Question

6. A mass of 135 g of an element is known to contain 3.01 × 1024 atoms. What is the element?

7. A young male adult takes in about 0.0005 m3 of fresh air during a normal breath. Fresh air contains approximately 21 % oxygen. Assuming that the pressure in the lungs is 100000 Pa, and air is an ideal gas at a temperature of 310 K, what is the number of oxygen molecules in a normal breath?

Answer 1

Mass of an atom is

Mass of an atom is m = 135 g/3.01 times 10^24 = 4.48 times 10^-23 g So, the mass of 1 mole of element is m N_A = 4.48 times 10^-23 times 6.02 times 10^23 g/mole = 26.97 g/mole From the periodic table we see that Aluminum has a mass of 27.98 g/mole. So, the given element is Aluminum. From ideal gas equation, we get the number of moles of air as pV = n RT n = pV/RT = 10^5 times 5 times 10^-4/8.314 times 310 = 0.0194 So, number of molecules of air are n N_A = 0.0194 times 6.02 times 10^23 = 1.17 times 10^22 Since air has 21% of oxygen molecules, we get no. of oxygen molecules as = 2.46 times 10^21

So, the mass of 1 mole of element is

$mN_A=4.48\times 10^{-23}\times 6.02\times10^{23}\;g/mole=26.97\;g/mole$

From the periodic table we see that Aluminum has a mass of 27.98 g/mole. So, the given element is Aluminum.

From ideal gas equation, we get the number of moles of air as

$pV=nRT\\ \therefore n=\frac{pV}{RT}=\frac{10^5\times 5\times10^{-4}}{8.314\times 310}=0.0194$

So, number of molecules of air are

$nN_A=0.0194\times6.02\times10^{23}=1.17\times 10^{22}$

Since air has 21% of oxygen molecules, we get no. of oxygen molecules as

$0.21\times1.17\times 10^{22}=2.46\times10^{21}$