Question

The mass of oxygen at standard conditions of temperature and pressure is 1.43g and that of a litre of $\text{S}{{\text{O}}_{2}}$is 2.857g.
(i) how many molecules of each gas are there in this volume?
(ii) what is the mass in grams of a single molecule of each gas?
(iii) what are the molecular masses of $\text{S}{{\text{O}}_{2}}$and ${{\text{O}}_{2}}$respectively?

Answer 1

Hint: We can solve it by using the concept of Avogadro’s law which states that the each substance at standard conditions of pressure(1 atm ) and temperature(25$^{\circ }\text{C}$) occupies 22.4 L of the volume and contains Avogadro number of molecules i.e. $\text{6}\text{.023}\times {{10}^{23}}$. So, from this we can easily solve all the given parts of numerical.

Complete step by step answer:
Oxygen and Sulphur are the elements which belong to the 16th group of the periodic table and are non-metals and have the atomic number as 8 and 16 respectively and the molecular mass as 16 and 32 respectively.
Since there are standard conditions of temperature and pressure, that means it follows the Avogadro’s law which states that 1 mol each atom, element or the compound contains the Avogadro’s number of molecules i.e. $\text{6}\text{.023}\times {{10}^{23}}$ at standard conditions of 1atm pressure and 25$^{\circ }\text{C}$ temperature and each atom, element or the compound occupies the volume of 22.4 L at that standard conditions of temperature and pressure.
Now, consider the (i) part, we know that;
1mole of each gas(i.e. Both $\text{S}{{\text{O}}_{2}}$and ${{\text{O}}_{2}}$)at STP occupies= 22.4L
Since, we know that the gas taken is 1 litre(given), then;
1 L of each gas(i.e. Both $\text{S}{{\text{O}}_{2}}$and ${{\text{O}}_{2}}$)at STP occupies=$\dfrac{1}{22.4}$ L
$\text{6}\text{.023}\times {{10}^{23}}$molecules each gas(i.e. Both $\text{S}{{\text{O}}_{2}}$and ${{\text{O}}_{2}}$)at STP occupies=$\dfrac{\text{6}\text{.023}\times {{10}^{23}}}{22.4}$L
= 0.2688$\times {{10}^{23}}$L
= 2.68$\times {{10}^{22}}$L
(ii) we know that,
22.4 L of gas${{\text{O}}_{2}}$contains = $\text{6}\text{.023}\times {{10}^{23}}$ molecules at STP
1 L of gas ${{\text{O}}_{2}}$will contain=$\dfrac{\text{6}\text{.023}\times {{10}^{23}}}{22.4}$molecules at STP
= 0.2688$\times {{10}^{23}}$ molecules at STP
= 2.68$\times {{10}^{22}}$ molecules at STP
So, now 2.68$\times {{10}^{22}}$ molecules of ${{\text{O}}_{2}}$ at STP has= 1.43 g of mass of ${{\text{O}}_{2}}$
1 molecule at STP has =$\dfrac{1.43}{2.68~\times {{10}^{22}}}$g of mass of ${{\text{O}}_{2}}$
= 5.32 $\times {{10}^{-22}}$g of mass of ${{\text{O}}_{2}}$
Similarly, 22.4 L of gas$\text{S}{{\text{O}}_{2}}$contains = $\text{6}\text{.023}\times {{10}^{23}}$ molecules at STP
1 L of gas $\text{S}{{\text{O}}_{2}}$will contain=$\dfrac{\text{6}\text{.023}\times {{10}^{23}}}{22.4}$molecules at STP
 = 0.2688$\times {{10}^{23}}$ molecules at STP
 = 2.68$\times {{10}^{22}}$ molecules at STP
So, now 2.68$\times {{10}^{22}}$ molecules of $\text{S}{{\text{O}}_{2}}$ at STP has= 2.857 g of mass of $\text{S}{{\text{O}}_{2}}$
1 molecule at STP has =$\dfrac{2.857}{2.68~\times {{10}^{22}}}$g of mass of $\text{S}{{\text{O}}_{2}}$
= 1.06 $\times {{10}^{-22}}$g of mass of $\text{S}{{\text{O}}_{2}}$
 (iii) as we know that,
1L of ${{\text{O}}_{2}}$at STP contains= 1.43 g of ${{\text{O}}_{2}}$ (given)
22.4L of ${{\text{O}}_{2}}$at STP will contains= 1.43×22.4 g of ${{\text{O}}_{2}}$
= 32 g of ${{\text{O}}_{2}}$
Similarly, 1L of $\text{S}{{\text{O}}_{2}}$at STP contains= 2.857 g of $\text{S}{{\text{O}}_{2}}$ (given)
22.4L of ${{\text{O}}_{2}}$at STP will contains= 2.857×22.4 g of $\text{S}{{\text{O}}_{2}}$
= 64 g of $\text{S}{{\text{O}}_{2}}$

Note: Avogadro’s law is applicable only at the conditions of standard temperature and pressure and if the pressure and temperature are constant , then on increasing the amount of the gas, its volume also increases and vice-versa.