Question

Calculate the atomicity of chlorine, if 35.5g of it occupies 11,200 cm³ at S.T.P.​

Answer 1

Solution:-

Here we have been given,

• 35.5 g of chlorine occupies 11,200 cm³ at S.T.P.

We have to calculate the atomicity of chlorine.

As we know 1 mole of any particular substance is equal to 1 g molecular mass of it.

i.e., 1 mole of substance = 1 g molecular mass of it

___

1 gram of chlorine occupies how much ?

==> 11200cm³

Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.

==> 22400 × 35.5 / 11200

==> 22400 × 355 / 11200 × 10

==> 224 × 355 / 112 × 10

==> 224 × 71 / 112 × 2

==> 112 × 71 / 112

==> 71

Thus, gram molecular mass would be 71g .

Now at last we would calculating the atomicity by using this formula:

==> Atomicity = Molecular mass / Atomic mass

Again given,

• Molecular mass = 71g

• Atomic mass = 35.5 g

Applying the required values:

==> 71 / 35.5

Solving now,

==> 71 × 10 / 355

==> 71 × 2 / 71

==> 2

Therefore, atomicity of chlorine is 2

Additional information:-

• The relative molecular mass of a gas is twice its vapour density.

• The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

• The molecular mass of a substance is expressed in grams is called gram molecular mass or molar mass.

• The molar volume of a gas can be defined as the volume occupied by one mile of a gas at S.T.P.

Answer 2

$\huge\tt{\fbox{\purple{Answer}}}$

Given-

35.5 g of chlorine occupies 11,200 cm³ at S.T.P.

35.5 g of chlorine occupies 11,200 cm³ at S.T.P. We have to calculate the atomicity of chlorine.

35.5 g of chlorine occupies 11,200 cm³ at S.T.P. We have to calculate the atomicity of chlorine. As we know 1 mole of any particular substance is equal to 1 g molecular mass of it.

i.e., 1 mole of substance = 1 g molecular mass of it

1 gram of chlorine occupies how much ?

1 gram of chlorine occupies how much ? ==> 11200cm³

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10 ==> 224 × 71 / 112 × 2

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10 ==> 224 × 71 / 112 × 2 ==> 112 × 71 / 112

1 gram of chlorine occupies how much ? ==> 11200cm³ Calculating grams of chlorine that will occupy 22,400 cm³ at S.T.P.==> 22400 × 35.5 / 11200 ==> 22400 × 355 / 11200 × 10 ==> 224 × 355 / 112 × 10 ==> 224 × 71 / 112 × 2 ==> 112 × 71 / 112 ==> 71

-------------------------------------------------

Thus, gram molecular mass would be 71g .

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula:

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic mass

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values:

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,==> 71 × 10 / 355

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,==> 71 × 10 / 355==> 71 × 2 / 71

Thus, gram molecular mass would be 71g .Now at last we would calculating the atomicity by using this formula: ==> Atomicity = Molecular mass / Atomic massAgain given,• Molecular mass = 71g • Atomic mass = 35.5 g Applying the required values: ==> 71 / 35.5 Solving now,==> 71 × 10 / 355==> 71 × 2 / 71 ==> 2

=> atomicity of chlorine is 2