calculate density of CH4 gas at 0C 700 mass Hg pressure.Molecular mass of CH4 is16

If I correct the typos to make the problem read

Calculate the density of CH4 gas at 0C and 700 mm Hg pressure.
The reworked universal gas law of PV = nRT becomes P*molar mass = density*RT
P is 700/760
T is 273K for 0C.
R is 0.08206 L*atm/mol*K
M = 16
Solve for density.

Density=PM/RT

760 mm Hg = 1 Atm
0 C= 273 K
Density=1(12+4)/0.08206*273)=0.71 g/L

To calculate the density of CH4 gas at a given temperature and pressure, we can use the Ideal Gas Law equation:

PV = nRT

Where:
P = pressure (in units of force per unit area)
V = volume (in units of cubic meters)
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin. The conversion formula is:
T(K) = T(C) + 273.15

So, the temperature in Kelvin would be:
T(K) = 0 + 273.15 = 273.15 K

Next, we convert the given pressure from mass Hg (mmHg) to SI unit Pascal (Pa). The conversion factor is:
1 mmHg = 133.322 Pa

So, the pressure in Pascal (Pa) would be:
P(Pa) = 700 mmHg * 133.322 Pa = 93,325.4 Pa

Now, we rearrange the Ideal Gas Law equation to solve for density (ρ), which is mass per unit volume (kg/m^3):
ρ = (PM) / (RT)

Where:
P = pressure (in Pascal)
M = molar mass of the gas (in kg/mol)
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)

The molar mass of CH4 is 16 g/mol, which is 0.016 kg/mol.

Substituting the values into the equation:
ρ = (93,325.4 Pa * 0.016 kg/mol) / (8.314 J/(mol·K) * 273.15 K)

Simplifying:
ρ = 601.95 kg/m^3

Therefore, the density of CH4 gas at 0°C and 700 mmHg pressure is approximately 601.95 kg/m^3.