701.28 grams of SO2 is held at 3606.96 torr and 603.2oC. What is the volume of the container

can some one answer this?

165.92L

Explanation:

First, let us calculate the number of mole SO2 in the container. This is illustrated below:

Molar Mass of SO2 = 32 + (16x2) = 32 + 32 = 64g/mol

Mass of SO2 = 701.28g

Number of mole = Mass / Molar Mass

Number of mole of SO2 = 701.28/64 = 10.96moles

Now we can easily calculate the volume of the container by calculating the volume of SO2 as illustrated below:

Data obtained from the question include:

P = 3606.96 torr

Recall: 760torr = 1atm

Therefore, 3606.96 torr = 3606.96/760 = 4.746atm

T = 603.2°C = 603.2 + 273 = 876.2K

R = 0.082atm. L/Kmol

n = 10.96moles

V = ?

Using the ideal gas equation: PV = nRT, the volume can be obtain as follow:

PV = nRT

V = nRT/P

V = (10.96 x 0.082 x 876.2) / 4.746

V = 165.92L

Therefore, the volume of the container is 165.92L