How can i calculate the volume of 40.6g of F2 AT STP???

Answer 1

convert ftom g to mole and then from moles too liters.

1 Mole = approx 38 g so 40.6 g * 1 mole / 38 g = 1.07 moles

and since 1 Mole @ STP = 22.4 liters

then 1.07 moles * 22.4 L / mole = 23.9 liters

Answer 2

Use the ideal gas equation PV=nRT

Convert 40.6 g of F2 to moles using the molar mass, that gives you the value of n. STP means standard temperature and pressure. Standart temp is 273 K, and standard pressure is 1atm. Use the ideal gas constant of 0.08206 l atm/mol K.

solve the equation for V

Answer 3

F2 has a molecular mass of 38g (2*mass of F=19)
so 40.6g correspond to =1.06842 moles
STP correspond at 10^5 Pa and 273.15°K

so pV= nRT

V= nRT/p = 1.06842*8.31*273.15/10^5 =0.02425m^3= 24.25 L

Answer 4

I hope by STP you mean normal temperature ( 0 degrees C ) and atmospheric pressure.

Remember that 1 gm-mole of every ideal gas occupies 22.4 litre at STP. Molecular wt. of F is 38. Hence, 38 gm occupy 22.4 litre. So 40.6 g will occupy

40.6 x 22.4 / 38 = 23.93 litre

Answer 5

For this we need the equation PV = nRT
P = 101.3 kPa
V = ?
R = 8.315
T = 273 K
n = m / M
= 40.6 / 38
= 1.068 mol

V = nRT / P
= 1.068*8.315*273 / 101.3
= 23.9 L