Which sample of gas at STP has the same number of molecules as 6 liters of Cl2(g) at STP?

The correct answer is option 2  (6 liters of N2)

Explanation:

The complete question

Which sample of gas at STP has the same number  of molecules as 6 liters of Cl2(g) at STP?

(1) 3 liters of O2(g)

(2) 6 liters of N2(g)

(3) 3 moles of O2(g)

(4) 6 moles of N2(g)

Step 1: Data given

Volume = 6 L

STP = 1 atm and 273 K

Step 2: Calculate moles of Cl2

p*V = n*R*T

n = (p*V)/ (R*T)

⇒with n = the number of moles Cl2 = TO BE DETERMINED

⇒with V = the volume of Cl2 = 6.0 L

⇒with p = the pressure of Cl2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles Cl2

This means we need 0.2678 moles of a gas at STP

Option 3 has 3 moles of O2 ⇒ Not the same number of molecules

Option 4 has 6 moles of N2  ⇒ Not the same number of molecules

For 3 liters of O2 we'll have:

⇒with n = the number of moles O2 = TO BE DETERMINED

⇒with V = the volume of O2 = 3.0 L

⇒with p = the pressure of O2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 3.0 ) (0.08206 * 273)

n = 0,1339 moles ⇒ Not the same number of molecules

For 6 liters of N2 we'll have

⇒with n = the number of moles N2 = TO BE DETERMINED

⇒with V = the volume of N2 = 6.0 L

⇒with p = the pressure of N2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles N2 ⇒ The same number of molecules

The correct answer is option 2  (6 liters of N2)

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