What is the molecular weight of a gas if a 15.0g sample has a pressure of 836 mm Hg at 25.0 degrees C in a 2.00 L flask?

a)167
b)1.35
c)176
d)11.1
e)none of the above

Please explain... I need to understand how to do this. Thanks.

Use PV = nRT to solve for n.

Then n = grams/molar mass.

I'm still confused... I know that:

P=1.1 atm
V=2.00 L
n=?
R= I think .0821
and
T= ?

I have no idea how to change 15.0g to moles, since I do not know what the element is to look at the atomic mass--I know that one mole is avagadro's number, but I'm still really not sure what to do here.

Please help...

T- 25+273 = 298k

n=PV/RT

n= 1.1 atm*2.00 l/ (0.0821 (L*atm/ mol*k) *298k)
n=0.0899mol
MW= 5.0g/0.0899mol

=167amu

To determine the molecular weight of a gas using the given information, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

First, we need to convert the given pressure from mm Hg to atm. Since 1 atm equals 760 mm Hg, we can use the following conversion factor:

1 atm = 760 mm Hg

836 mm Hg × (1 atm/760 mm Hg) = 1.097 atm (rounded to three decimal places)

Next, we need to convert the temperature from degrees Celsius to Kelvin. To do this, we add 273.15 to the given temperature:

25.0 degrees C + 273.15 = 298.15 K

Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):

n = (PV) / (RT)

n = (1.097 atm × 2.00 L) / (0.0821 L·atm/mol·K × 298.15 K)

n ≈ 0.0905 moles (rounded to four decimal places)

Finally, we can calculate the molecular weight by dividing the given mass (15.0 g) by the number of moles:

Molecular weight = mass / moles

Molecular weight = 15.0 g / 0.0905 moles

Molecular weight ≈ 165.75 g/mol (rounded to two decimal places)

Comparing the calculated molecular weight to the options given, we find that the closest option is 167 g/mol (option a). Therefore, the correct answer is (a) 167.