Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 5 (Q.No. 46)
46.
The total number of atoms in 8.5 gm of NH3 is __________ x 1023.
Discussion:
9 comments Page 1 of 1.
Nader said:
8 years ago
Answer [A] is definitely the right answer.
n = M / Mwt.
n = 8.5 / 17 = 0.5 mole.
Thus, the number of "MOLECULES" = 0.5 * 6.023*10^23 = 3.01 MOLECULES.
Therefore, the number of Atoms = 3.01 * 4 = 9.03.
Since, each NH3 molecule has 4 Atoms.
n = M / Mwt.
n = 8.5 / 17 = 0.5 mole.
Thus, the number of "MOLECULES" = 0.5 * 6.023*10^23 = 3.01 MOLECULES.
Therefore, the number of Atoms = 3.01 * 4 = 9.03.
Since, each NH3 molecule has 4 Atoms.
Rampravesh yadav said:
5 years ago
@All.
Moles of NH3 will be =0.5
And then we know that @
1 mole have 6.023*10^23
Then we have multiplied by 4 in to 3.023*10^23 then we find.
So {12.092* 10^23} is the answer.
Moles of NH3 will be =0.5
And then we know that @
1 mole have 6.023*10^23
Then we have multiplied by 4 in to 3.023*10^23 then we find.
So {12.092* 10^23} is the answer.
Suraj Awathe said:
5 years ago
@All.
8.5gm NH3 =0.5 Moles of NH3.
One mole contains 6.022 *10^23.atoms also known as Avogadro's number or Avogadro's constant.
Hence 0.5moles= 3.01*10^23
8.5gm NH3 =0.5 Moles of NH3.
One mole contains 6.022 *10^23.atoms also known as Avogadro's number or Avogadro's constant.
Hence 0.5moles= 3.01*10^23
Kishan nariya said:
9 years ago
Answer is B
Because 1mol of NH3 > 17 gm.
NH3 > 6.023 * 10^23.
8.5 gm NH3 > 3.01 * 10^23.
Because 1mol of NH3 > 17 gm.
NH3 > 6.023 * 10^23.
8.5 gm NH3 > 3.01 * 10^23.
(1)
Kishan nariya said:
9 years ago
Please give correct answer of the question. Whether it is option A or B.
Hanadi said:
8 years ago
How you calculate 3.01*4=9.03?
3.01*4=12.04.
Please tell me how?
3.01*4=12.04.
Please tell me how?
(1)
Yaduvanshi said:
5 years ago
@Rampravesh yadav.
You are absolutely right. Thanks.
You are absolutely right. Thanks.
TANMAY SHAH said:
2 years ago
I also think 3.01 is the Right Answer.
(2)
Anup said:
6 years ago
According to me, it's option C.
(1)
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