If a sample of 12.0 grams of fluorine gas at 45.00C has a pressure of 0.544 atm, what is the volume of the container? R = 0.0821 L atm/mol K

15.1 L

4.29 L

30.3 L

2.14 L

Volume of the container is 30.3 L.

Explanation:

Given: Mass = 12 g

Temperature = = (45 + 273) K = 318 K

Pressure = 0.544 atm

R = gas constant = 0.0821 L atm/mol K

Moles is the mass of substance divided by its molar mass.

Hence, moles of fluorine (molar mass = 18.99 g/mol) are as follows.

Formula used to calculate the volume of container is as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant

T = temperature

Substitute the value into above formula as follows.

Thus, we can conclude that volume of the container is 30.3 L.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml