Question

The number of molecules present in 200 g of ${\text{Na}}_2{\text{CO}}_3$ is

• A. $3.19\times {10}^{23}$
• B. $1.81\times {10}^{24}$
• C. $1.14\times {10}^{24}$
• D. $0.31\times {10}^{-23}$

Answer 1

The correct option is C

$1.14\times {10}^{24}$

Given mass = 200 g
Avogadro number = $6.022\times {10}^{23}$
Molar mass of ${\text{Na}}_2{\text{CO}}_3$ =
2 $\times$ Atomic mass of sodium + 1 $\times$ Atomic mass of carbon + 3 $\times$ Atomic mass of oxygen
= $2\times 23+12+3\times 16$
= 106

Number of molecules of sodium carbonate
= Number of moles $\times$ Avogadro number ...(i)
Also, number of moles of sodium carbonate = $\frac{\text{Mass}}{\text{Molar mass}}$ ...(ii)
Inserting equation (ii) in equation (i), we get
Number of molecules of sodium carbonate = $\frac{\text{Mass}}{\text{Molar mass}}\times \text{Avogadro number}$ ...(iii)

Inserting values of mass, Avogadro number and molar mass in equation (iii) we get
Number of molecules of sodium carbonate = $\frac{200}{106}\times 6.022\times {10}^{23}$
= $1.88\times 6.022\times {10}^{23}$
$\therefore$ Number of molecules of sodium carbonate
= $1.13\times {10}^{24}$