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The volume of 17 gm. NH3 gas at N.T.P. is (22.4/44.8/11.2/88.8)liter

The volume of 17 gm. NH3 gas at N.T.P. is (22.4/44.8/11.2/88.8)liter

Grade:9

3 Answers

Ritik
55 Points
6 years ago
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
12 Points
6 years ago
 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
19 Points
6 years ago
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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