1) A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of

Question

1) A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 8.7 atm ?

2) An experiment shows that a 116 mL gas sample has a mass of 0.168 g at a pressure of 694 mmHgand a temperature of 31 C. What is the molar mass of the gas?

3) A sample of gas has a mass of 0.565 g . Its volume is 119 mL at a temperature of 85 C and a pressure of 758 mmHg. Find the molar mass of the gas.

Please answer all parts.

Explanation / Answer

1)

m = 2 g of O2 per m = 98 g of He

P of O2

Pt = 8.7 atm

change all to mol

mol of O" = mass/M" = 2/32 = 0.0625

mol of He = mass/MW = 98/4 = 24.5

total mol 24.5+0.0625 = 25.125mol

mol frac of O2 = 0.0625/25.125 = 0.00248

then

Partial P of O2 = xO2 * PT = 0.00248*8.7 = 0.02157 atm of O2

2)

PV = nRT

n = mass/MW

then

PV = mass/MW*RT

solve for MW

MW = mass*RT/(PV)

P = 694/760 = 0.913157atm

MW = (0.168)(0.082)(31+273)/(0.913157*0.116)

MW = 39.536 g /mol (probably argon)

3)

m = 0.565

V = 119 ml = 0.119 L T = 85 °C = 85+273 = 358 K

P = 758 mm Hg = 758/760 atm = 0.9973 atm

then

PV = nRT

n = m/MW so

PV = m/MW * RT

MW = m*RT/(PV)

MW = (0.565 *0.082)(358 )/(0.9973*0.119) = 139.756 g/mol

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.