How many moles of carbon dioxide are contained in 237.8 l at stp?

STP = Standard Temperature and pressure 
T = 0°C or 273.15 K 
P = 1 atm 

Given V = 237.8 L 
and we know gas constant, R = 0.082 L atm / K mol 

PV = nRT 
n = PV/RT = (1 x 237.8) / (0.082 x 273.15) = 10.62 mol 

 
Quicker way is to do as "Roger the Mole" did above 

At STP 1 mol of any gas occupies 22.4 L 
then moles of CO2 = (237.8 L) / (22.4 L/mol) = 10.62 mol

Ithink the answer is a or d.