Number of neutrons present in 1-7 g of ammonia is-0-8N-A0-7N-A0-652N-A0-52N-A

Question

Toppr Admin · Accepted Answer

Let us consider the problem-For- moles of NH3-1-717-0-1Let the Neutrons in 1 molecules of NH3-7-3-xD7-0-7Hence-Number of Neurons -0-1-xD7-NA-xD7-7The correct answer is 0-7NA

Number of neutrons present in $1.7$ g of ammonia is:
$0.8 N_{A}$
$0.7 N_{A}$
$0.652 N_{A}$
$0.52 N_{A}$

Let us consider the problem.
For, moles of $N H_{3} = \frac{1.7}{17} = 0.1$
Let the Neutrons in 1 molecules of $N H_{3} = 7 + 3 \times 0 = 7$
Hence,
Number of Neurons $= 0.1 \times N_{A} \times 7$
The correct answer is $0.7 N_{A}$

Number of neutrons present in 1.7 g of ammonia is:0.8NA0.7NA0.652NA0.52NA

Let us consider the problem.For, moles of NH3=1.717=0.1Let the Neutrons in 1 molecules of NH3=7+3×0=7Hence,Number of Neurons =0.1×NA×7The correct answer is 0.7NA

Number of neutrons present in $1.7$ g of ammonia is:
$0.8 N_{A}$
$0.7 N_{A}$
$0.652 N_{A}$
$0.52 N_{A}$

Let us consider the problem.
For, moles of $N H_{3} = \frac{1.7}{17} = 0.1$
Let the Neutrons in 1 molecules of $N H_{3} = 7 + 3 \times 0 = 7$
Hence,
Number of Neurons $= 0.1 \times N_{A} \times 7$
The correct answer is $0.7 N_{A}$