Question

The number of atoms in 1.7 g of NH3 is​

Answer 1

Answer:

1.7 g of the ammonia contains (6.022 × 10²³) × 1.7/17 atoms.

Explanation:

now calculate using calculator

Answer 2

Answer:

The number of atoms present in 1.7g of NH₃ is equal to 2.4092×10²³ atoms.

Explanation:

Given, the mass of ammonia = 1.7g

The molecular mass of ammonia = 14+3(1)=17gmol⁻¹

We know that one mole of ammonia contains 6.023×10²³ molecules of ammonia.

One molecule of ammonia contains =4 atoms

17g of ammonia contains = 1mole

So, 1.7g of ammonia will have $=\frac{1.7}{17}=0.1mole$

1 mole of NH₃ contains = 6.023×10²³ molecules

0.1 mole of NH₃ contains $=6.023*10^{23}*0.1=0.6023*10^{23}\;molecules$  

Now,  One molecule of ammonia contains =4 atoms

0.6023×10²³ molecules of NH₃ have atoms $=0.6023*10^{23}*4$

                                                                        $=2.4092*10^{23}$ atoms

Therefore, the number of atoms is equal to  2.4092×10²³ present in 1.7g of ammonia.