Question

The molecules present in 5.6 Lof sulphur dioxide at STP is​

Answer 1

Given :

Volume of sulphur dioxide at STP = 5.6 L

To Find :

No of molecular present in the compound.

Solution :

❒ First of all we need to find number of moles of sulphur dioxide.

We know that, Volume of 1 mole of compound at STP is equal to 22.4 L.

Therefore we can say that

$\bf\longrightarrow\:No.\:of\:moles=\dfrac{Given\:volume}{22.4}$

By substituting the given values,

➝ n = V / 22.4

➝ n = 5.6 / 22.4

➝ n = 1 / 4

➝ n = 0.25

❒ One mole of compound contains 6.022 × 10²³ molecules$.$

Therefore, number of molecules present in 0.25 moles of compound is given by

➝ N = n × 6.022 × 10²³

➝ N = 0.25 × 6.022 × 10²³

➝ N = 1.5 × 10²³ molecules

Answer 2

Given :

Volume of sulphur dioxide at STP = 5.6 L

To Find :

No of molecular present in the compound.

Solution :

First of all we need to find number of moles of sulphur dioxide.

We know that, Volume of 1 mole of compound at STP is equal to 22.4 L.

Therefore we can say that

$\bf\longrightarrow\:No.\:of\:moles=\dfrac{Given\:volume}{22.4} \:$

By substituting the given values,

➙ n = V / 22.4

➙ n = 5.6 / 22.4

➙ n = 1 / 4

➙ n = 0.25

One mole of compound contains 6.022 × 10²³ molecules..

Therefore, number of molecules present in 0.25 moles of compound is given by

➙ N = n × 6.022 × 10²³

➙ N = 0.25 × 6.022 × 10²³

➙ N = 1.5 × 10²³ molecules.

∴ The molecules present in 5.6L of sulphur dioxide at STP is 1.5 × 10²³.