If 0.80 g of a gas occupies 295 mL at 25 degree C and 680 mm Hg of pressure, wha

Question

If 0.80 g of a gas occupies 295 mL at 25 degree C and 680 mm Hg of pressure, what v the molar mass of the gas? A sample of nitrogen occupies a volume of 300 mL at 30 degree C and 700 mm Hg of pressure. What will be its volume at STP? Consider Figure 10.1. If the height of the mercury column in the leveling bulb is 30 mm greater than that in the gas buret and atmospheric pressure is 670 mm what is the pressure on the gas trapped in the buret? Calculate the density of N_2 at STP (a) using the ideal gas law and (b) using the molar volume and molar mass of N_2. How do the densities compare?

Explanation / Answer

6)

According to ideal gas equation

PV = nRT

Where P = pressure of gas ; V = volume of gas ; n = number of moles = mass in grams/Molar mass= w/M

R = universal gas constant     ; T = Temparature

P = 680 mm Hg = 0.89 atm

V = 250 ml = 0.250 L

Mass in grams = 0.80 g

T = 250C = 298 K

So PV = wRT /M                  ; M = wRT / PV

M = (0.80 g x 0.0821 L atm mol-1K-1 x 298 K) / 0.89 atm x 0.250 L = 87.96 = 88.0 g/mol

Molar mass = 88.0 g/mol

7)

PV = nRT

Where P = pressure of gas ; V = volume of gas ; n = number of moles = mass in grams/Molar mass= w/M

R = universal gas constant     ; T = Temparature

P =700 mm Hg = 0.92 atm

V = 300 ml = 0.300 L

T = 300C = 300 K

n = PV /RT = (0.92 atm x  0.300 L) / 0.0821 L atm mol-1K-1 x 300 = 0.011 mol

According to Avagadro's law at STP molar volume of a gas is 22.4 L

1 mole of gas = 22.4 L

So at STP 0.011 mol gas occupies = 22.4 L x 0.011 = 0.2464 L

9)

Density = mass in g / Volume in L

According to Avagadro's law at STP molar volume of a gas is 22.4 L

1 mole of N2 = 28 g

Density = 28 g / 22.4 L = 1.25 g/L = 1.25 x 10-3 g/cm3

According to ideal gas equation

PV = nRT

For 1 mole PV = wRT / M

Or w/V = MP /RT

Where w/V = Density = mass in g / Volume in L

Standard temparature T = 273.15 K , P = 1 atm , R = 0.0821 L atm mol-1K-1 , M = 28 g/mol

w/V = Density = (28 g/mol x 1 atm) /   0.0821 L atm mol-1K-1 x 273.15 K = 1.25 g /L = 1.25 x 10-3 g/cm3

So both are equal

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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