Question

(8) Calculate the no. of moles present in 0.0056 m
Ar and 224 cm3 of He.
(2 mark
do humanbite nencilis of weig​

Answer 1

Given:

The following quantities of Ar and He

Ar  : 0.0056 m³

He :  224 cm³

To find :

The number of moles of present of each gas

Concept:

As per Avogadro's law at NTP  22.4 litre of a gas = 1 gram mole

Solution:

1. Argon (Ar)

Given volume V= 0.0056 m³

=0.056 × 1000 liters  ( 1 m³=1000 liters)

V = 56 liters

$No. \,of\, moles = \frac{ volume}{22.4}\\\\= \frac {56}{22.4}\\\\\boxed{No. \,of\, moles\, of\, Ar= 2.5 }$

2. Helium (He)

Given volume V= 224 cm³

$V= \frac{224}{1000} =0.224\, Liters\, (\, \because 1 \,cm^3=\frac{1}{1000} Liter)\\\\Thus \\\\\boxed{No.\, of\, moles = \frac{0.224}{22.4} = 0.01 }$