Question

What is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 degrees celcius and 744.0 mm Hg?

Answer 1

The molar mass of a gas with a density of 0.00249 g/mL at 20.0 °C and 744.0 mm Hg is calculated to be approximately 62.48 g/mol using the ideal gas law and the density formula.

Step-by-step explanation:

To calculate the molar mass of a gas given its density, we can use the ideal gas law in combination with the density formula.
One formula to use for the molar mass (M) when given the gas density (ρ), pressure (P), and temperature (T) is: M = dRT/P, where R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin. We must first convert temperature to Kelvin by adding 273.15 to the Celsius temperature, and the pressure from mm Hg to atm using the conversion factor (1 atm = 760 mm Hg).

Convert 20.0 °C to Kelvin: T = 20.0 + 273.15

= 293.15 K
Convert pressure from mm Hg to atm:

P = 744.0 mm Hg × (1 atm / 760 mm Hg)

= 0.979 atm

Next, we use the density given in g/mL and convert to g/L: density

= 0.00249 g/mL × 1000 mL/L

= 2.49 g/L.

Now, apply the formula to get the molar mass of the gas:
M = (dRT)/P

= (2.49 g/L × 0.0821 L·atm/K·mol × 293.15 K)/0.979 atm

= 62.48 g/mol

Answer 2

61.24 is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 degrees celcius and 744.0 mm Hg.

Step-by-step explanation:

given that:

density = 0.00249 g/ml (
`(mass)/(volume)`) or 2.49 grams/litre

P = 744 mm Hg OR 0.978 atm

T = 20 Degrees or 293.15 Kelvin

R = 0.08206 Litre atm/mole K

molar mass =?

Formula used/

PV = nRT equation 1

here n is number of moles:

n =
`(mass)/(atomic mass)`

putting the value of n and value of density in the equation 1:

PV =
`(mass)/(molar mass)` x RT

molar mass =
`(mass)/(volume)` x
`(RT)/(P)`

= density x
`(RT)/(P)`

=
`(2.49 x 0.08276 x 293.15)/(0.978)`

= 61.24 is the molar mass of the gas.