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What is the number of atoms present in 4.25 L of ammonia gas at 373 K? Given, molar volume (Volume of 1 mol) of NH 3 at 373 K is 42.5 Lmol -1

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kinsondols186

Answer:

There are 6.023*10^22 atoms present in 4.25L of NH3 at 373K

Explanation:

Given:

1 mol of NH3 at 373K= 42.5L/mol

Thus, No. of mol of NH3 at 373K= Given Volume/Molar volume

=4.25L/42.5L/mol

=0.1 mol

So, 1 mole of NH3= 6.023*10^23 atoms

0.1 mol of NH3= 6.023*10^23*0.1 atoms

= 6.023*10^22 atoms

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