Question

What is the density (in g/L) of a gas with a molar mass of 16.01 g/mol at 1.75 ATM and 337 K?

Answer 1

You can use this formula to solve for density--> Density= PM/ RT, where P is pressure, M is molar mass, R is the gas constant and T is temperature. 

P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k

density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L

Answer 2

Answer : The density of a gas is, 1.013 g/L

Solution : Give,

Molar mass of gas = 16.01 g/mole

Pressure of gas = 1.75 atm

Temperature of gas = 337 K

Using ideal gas law equation,

$PV=nRT$

$PV=\frac{w}{M}\times RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

As we know that $Density=\frac{Mass}{Volume}$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of the gas

w = given mass of gas

M = molar mass of gas

R = gas constant = 0.0821 Latm/moleK

$\rho$ = density of gas

Now put all the given values in the above formula, we get the density of the gas.

$\rho=\frac{PM}{RT}$

$\rho=\frac{(1.75atm)\times (16.01g/mole)}{(0.0821Latm/moleK)\times (337K)}=1.013L$

Therefore, the density of a gas is, 1.013 g/L