What is the volume of 1.9 moles of chlorine gas (cl2) at 298 k and 1.3 atm?

Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol

The answer is 43 L if I am correct.

Hello!

For this exercise you have to use the Ideal Gas Law Equation, as follow:

PV= nRT

Where:

P= pressure ( in this example, atmosphere (atm) is the pressure unit)

V= volume (which is the value you want to calculate)

n= Mol (as reference to the mass of the gas Cl2)

R = Ideal Gas Constant (for this example is a number 0.0821 atm*L/mol*K)

T= Temperature in Kelvin (K)

So, for find the value of the volume, you just have to replace the values given in the equation as follow:

P= 1.3 atm

V = you don't know, is what you want to know.

n= 1.9 mol

R =0.0821 atmL/molK

T =298 K

Remember, never forget to put the units, because teacher can get down your score if you don't do it. So, back with the values to the equation:

13atm * V = 1.9mol * 0.0821 atmL/molK * 298K

Then, we have to solve the equation to find V value, so we has to send 1.3 atm to divide in the other side of the equation:

V = (1.9mol * 0.0821 atmL/molK * 298K)/ 1.3 atm

V = 35.76 L

And, that's it!

I hope this help you a lot, and good luck.

your answer would be “dry”