How many chlorine molecules are in 7.02 l of chlorine gas at stp?
a. 946.63 1023
b. 1.89 1023
c. 26.12 10-23
d. 5.21 10-21

The answer is:  [B]:  1.89 * 10²³  .


PV = nRT ;  is the equation for "STP" conditions; that is, the "ideal gas equation" .

                      i.e. when Pressure, "P" = 1.00 atm; 
                                    Temperature, "T" = 273 K;
 
                           R = the ideal gas constant = ((0.08206 L-atm/K-mol)
                           n = number of moles;

So, we plug in our known values:

(1.00atm) (7.02L) = ("n" mol) (0.08206 L-atm/K-mol) (273K); 

→ 7.02 L·atm = (? mol) (22.4 L·atm/mol) .

     (Note that the Molar Volume of a gas at STP is a constant using Avogadro's value of 22.4 L / mol.  1 mol of any ideal gas at STP occupies 22.4 L. An ideal gas takes the shape of its container)..

 → Divide EACH side of the equation by "(22.4 L·atm/mol)" ; 

→ 7.02 L·atm / = ("n" mol) (22.4 L·atm/mol) ;

→ 7.02 L·atm / (22.4 L·atm/mol) =
                     [("n"  mol) (22.4 L·atm/mol)]/(22.4 L·atm/mol);
 
to get:  

 → n = 0.313 mol ;

Note: 1 mole = 1 mol = 6.022 * 10²³  molecules.

  → So, 0.313 mol Cl₂ (g) molecules * [(6.022 * 10²³  molecules) / (1 mol)] =

  → [(0.313) * (6.022 *10²³) ] molecules of Cl₂ (g) ;

            →  =  1.88 * 10²³ molecules of Cl₂ g ; 

                   → which most closely corresponds with answer choice:
 
                                                       [B]:  1.89 * 10²³  .