Question

find number of moles in 630g of HNO3

who will know this they will answer step by step

​

Answer 1

Explanation:

Before we move forward with the solution of this question, let us discuss some important concepts.

1)Mole: a mole is a physical quantity which represents the amount of mass of the substance required to have a collective of 6.022×1023

atoms of the given substance. Mole is a widely used unit for calculating the amount of matter of a substance. One mole of any substance weighs about the same as the molecular mass of that substance.

Now moving back to the question, we have been given a sample of 6.3 grams of HNO3

. We know that the atomic weight of hydrogen is 1 amu, atomic weight of nitrogen is 14amu and the atomic weight of oxygen is 16amu. Hence the molecular weight of HNO3

can be calculated as:

Molecular weight of HNO3

= (no. of atoms of hydrogen) (atomic weight of hydrogen) +

(no. of atoms of nitrogen) (atomic weight of nitrogen) +

(no. of atoms of oxygen) (atomic weight of oxygen)

=(1)(1)+(1)(14)+(3)(16)

= 63 g/mol

Now, to calculate the number of moles of HNO3

present in the given sample, we use the formula:

Number of moles of HNO3=WeightofthegivensamplemolecularweightofHNO3

=WeightofthegivensamplemolecularweightofHNO3

=6.363

= 0.1 moles

Now, the number of atoms/molecules present in 1 mole of a substance is represented by a quantity known as Avogadro’s Number. This quantity is equivalent to 6.02×1023

atoms per mole of a substance. Since, we have 0.1 moles of HNO3

, the number of molecules of HNO3

present are:

Number of molecules = (Avogadro’s number) (number of moles of HNO3

=(6.02×1023)(0.1)

)

=0.602×1023

molecules

Hence, there are 0.602×1023

molecules of HNO3

present in a 6.3 gram sample of HNO3

.

HNO3

has 1 hydrogen, 1 nitrogen and 3 oxygen atoms. Hence the atomicity of these elements in the given compound are 1, 1, 3 respectively.

Hence, the total number of atoms of each element are:

No of atoms of H=0.1×1×6.02×1023=6.021022

No of atoms of N=0.1×1×6.02×1023=6.02×1022

No of atoms of O=0.1×1×6.02×1023=18.069×1022

Answer 2

Answer:

In 6.3 g of nitric acid, 6.022 X 10 22 atoms of hydrogen, 6.022 X 10 22 atoms of nitrogen and 18.06 X 10 22 atoms of oxygen are present.

Explanation:

Molecular Mass of HNO3 = (1 + 14 + 16X3)gm

= 63 gm

Therefore 63 gm HNO3=1 mole

1 g of nitric acid = 1/63 moles

⇒6.3 mole = 0.1 mole

As 1 mole = 6.022 X 1023 particles

⇒ 0.1 mole = 6.022 X 10 23 X 0.1 molecule

Hence 6.3 g of nitric acid contains 6.022 X 10 22 molecules

One molecule of HNO3 contain 1 hydrogen, 1 nitrogen and 3 oxygen atoms

Therefore,

In 6.3 g of nitric acid, 6.022 X 10 22 atoms of hydrogen, 6.022 X 10 22 atoms of nitrogen and 18.06 X 10 22 atoms of oxygen are present.