Question

Number of molecules present in 224 cc of a triatomic gas
at NTP is​

Answer 1

Answer:

0.009311

Explanation:

PV=nRT

n= PV/RT

  = (1)(0.224)/(0.0821)(293)

  = 0.009311

Answer 2

224 cc of a triatomic gas will contain 6.022 × 10²¹ molecules.

GIVEN

Volume of a triatomic gas = 224cc

TO FIND

Number of molecules of gas.

SOLUTION

We can simply solve the above problem as follows -

We know that,

1cc = 10⁻³ litre

Therefore,

244 cc = present in 224 × 10⁻³ Litres

To calculate the number of moles of this gas, we will apply the ideal gas equation -

PV = nRT

Where,

P = pressure of the gas = 1atm

Volume of the gas = 224 × 10⁻³ Litres

n= number of moles of gas

R = Ideal gas constant = 0.0821 atm.L/mol.k

T = Temperature = 273.5 k

Putting the values in the above formula, We get -

1 × 224× 10⁻³ = n× 0.0821×273.5

n = 224×10⁻³/0.0821×273.5

n = 224×10⁻³ / 22.4

n= 0.01 moles

According to stochiometry,

1 moles = 6.022 × 10²² molecules

So,

0.01 moles of gas = 6.022×10²³ × 0.01

= 6.022×10²¹ molecules

Hence, 224 cc of a triatomic gas will contain 6.022 × 10²¹ molecules.

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