Question

3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are

• A. $6.68\times {10}^{23}$
• B. $6.09\times {10}^{22}$
• C. $6.022\times {10}^{23}$
• D. $6.022\times {10}^{21}$

Answer 1

The correct option is A

$6.68\times {10}^{23}$

Explanation for correct answer:-

Option $\left(a\right)6.68\times {10}^{23}$

Step 1: Given data

• Mass number of Carbon = $12$
• Hydrogen = $1$
• The mass number of Oxygen = $16$.

Hence,

$$\begin{gathered} \mathrm{C}=12\times 12=144 \\ \mathrm{H}=1\times 22=22 \\ \mathrm{O}=16\times 11=176 \end{gathered}$$

Weight of Sucrose = $144+22+176=342g/mol$.

Step 2: Calculation of moles of sucrose

$342g$ mol of sucrose$\left({\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11}\right)$ contains = $11\times {\mathrm{N}}_{\mathrm{A}}\mathrm{atoms}\mathrm{of}\mathrm{oxygen},$

where${\mathrm{N}}_{\mathrm{A}}=6.023\times {10}^{23}$

So , $3.42g$ mol of sucrose contains =

$$\begin{gathered} \frac{3.42\times 11\times {\mathrm{N}}_{\mathrm{A}}}{342}\mathrm{atoms}\mathrm{of}\mathrm{oxygen} \\ =0.11{\mathrm{N}}_{\mathrm{A}} \end{gathered}$$

And, $18g$ water contains

$$\begin{gathered} =18g/(1\times 2+16)gmo{l}^{-1} \\ =18g/18gmo{l}^{-1} \\ =1moloxygen \end{gathered}$$

Step 3: Calculation of moles of water

A mol of water (${\mathrm{H}}_2\mathrm{O}$) contains $1\times {\mathrm{N}}_{\mathrm{A}}\mathrm{atom}\mathrm{of}\mathrm{oxygen}$

Step 4: Evaluating the Total number of oxygen atoms

Total number of oxygen atoms = $\mathrm{Number}\mathrm{of}\mathrm{oxygen}\mathrm{atoms}\mathrm{from}\mathrm{sucrose}+\mathrm{Number}\mathrm{of}\mathrm{oxygen}\mathrm{atoms}\mathrm{from}\mathrm{water}$

Total number of oxygen atoms

$$\begin{gathered} =0.11{\mathrm{N}}_{\mathrm{A}}+1.0{\mathrm{N}}_{\mathrm{A}} \\ =1.11{\mathrm{N}}_{\mathrm{A}} \end{gathered}$$

Step 5 : Calculating Number of oxygen atoms in solution

Number of oxygen atoms in solution $=1.11\times {\mathrm{N}}_{\mathrm{A}}$

Number of oxygen atoms in solution = $1.11\times 6.023\times {10}^{23}$

Number of oxygen atoms in solution = $6.68\times {10}^{23}$