Question

If 224ml of a triatomic gas has mass 1gm at 273k and 1 atm pressure, then the mass of one atom is

Answer 1

Moles of gas = 224ml/22400ml                        =0.01 moles  Moles = mass / molar mass  0.01 = 1 / molar mass  Thus, molar mass = 100g  Mass of one atom = molar mass/ NA  As the gas is triatomic so it will be Molar mass/ NA ×3 Putting the values in above formula 100/6.023×3 =5.53×10^-23

Answer 2

Answer: $5.53\times 10^{-23}g$

Explanation:

According to the ideal gas equation:'

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = 224 ml = 0.224 L   (1L=1000ml)

T= Temperature of the gas = 273 K

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= ?

$n=\frac{PV}{RT}=\frac{1\times 0.224}{0.0821\times 273}=0.01moles$

It is given that 0.01 moles weigh = 1 gram

Thus 1 mole will weigh =$\frac{1}{0.01}\times 1=100g$

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

As the gas is triatomic :

$3\times 6.023\times 10^{23}=18.07\times 10^{23}$ atoms will weigh 100 g

thus 1 atom will weigh=$\frac{100}{18.07\times 10^{23}}\times 1=5.53\times 10^{-23}g$

Thus mass of one atom is $5.53\times 10^{-23}g$