Question

ANSWER IS NEEDED URGENTLY!!!
what is a molar mass of 37.96 g of gas exerting a pressure of 3.29 atm on the walls of a 4.60 l container at 375 K

Answer 1

Answer:

The molar mass of the gas is 77.20 gm/k

Explanation:

the data given as :

p=3.29atm ,V=4.60 litres , T =37.5 mass of the gas =37.96 gram

using the ideal gas law will given the formula is

PV=nRT (where R=universal gas constant 0.08206 litre.atom /kW

also, number of moles is not given so applying the formula

n=mass ÷ molar mass of one mole of the gas .

n=m÷x(x molar mass ) (m mass given)

now putting the values in ideal gas law equation

PV= m÷xRT

3.29×4.60 =37.96litere x ×0.08206×375

15.134= 1168.1241÷15.13

x=77.20gm/mole

if all the unit's in the formula are put will get cancel only grams / mole will be there . molecular weight is given by gram/ mole

Answer 2

Answer:

The molar mass of 37.96 g of gas exerting a pressure of 3.29 atm on the walls of a 4.60 L container at 375 K is 77 g/mol.

Explanation:

According to the ideal gas law,

PV = nRT

Where,

P = pressure = 3.29 atm

V = volume = 4.60 L

R = universal gas constant = 0.082 L atm/K/mol

T = temperature = 375 K

n = number of moles

Substituting the given values in the above equation:

3.29 × 4.60 = n × 0.082 × 375

⇒n = $\frac{3.29*4.60}{0.082*375}$

     = 0.492 moles

We know that n = given mass ÷ molar mass

Given mass = 37.96 g

∴molar mass = given mass ÷ n

                     = $\frac{37.96}{0.492}$ = 76.53 g/mol ≈ 77 g/mol

Therefore the molar mass of 37.96 g of gas exerting a pressure of 3.29 atm on the walls of a 4.60 L container at 375 K is 77 g/mol.