Grade 9Physical Chemistry
The volume of 17 gm. NH3 gas at N.T.P. is (22.4/44.8/11.2/88.8)liter
3 Answers
Ritik
first we will calculate the mole of NH3
mole of NH3= 17/17=1mole (given weight/ molicular weight).
volume of 1 mole of gas at NTP =22.4L.
So volume of 1mole of NH3 at NTP=22.4L
homo sapiens
We have 17g of ammonia (NH3) over here ,
we need to find the volume at N.T.P. ,
so ,
no. of moles of 17 g of ammonia = given mass / molar mass ,
( n of NH3)
= 17g / 17 [ molar mass of NH3 = 17g]
||---------------||----------------------|| = 1 mol ------- eq. 1
now ,
moles of ammonia = volume / 22.4 {[ at N.T.P. the case value of volume is 22.4]}
(n of NH3)
1 mol = volume / 22.4 {[ from eq. no 1 ]}
volume = 22.4 /1
volume = 22.4 liters ( L)
formulae used: {1} moles = given mass/ molar mass
{2} moles= volume / 22.4
ans ; 22.4 L
Peer Anzar
We know the relation n= given mass÷ at.mass/mol.mass= N÷Na = V÷22.4 Given mass÷ Mol.mass= V÷22.417÷17=V÷22.4V=22.4
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