SATHEE: Solutions Question 147

Solutions Question 147

Question: The number of molecules in 4.25 g of ammonia is approximately [CBSE PMT 2002]

Options:

A) $0.5 \times 10^{23}$

B) $1.5 \times 10^{23}$

C) $3.5 \times 10^{23}$

D) $2.5 \times 10^{23}$

Show Answer

Answer:

Correct Answer: B

Solution:

$17, g m N H_{3}$ = 1 mole. Molecules of $N H_{3} = \frac{6.02 \times 10^{23} \times 4.25}{17} = 1.5 \times 10^{23}$

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