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Chemistry Question on Some basic concepts of chemistry

The total number of electrons in 18 ml18\, ml of water (density =1 g ml−1= 1\, g \,ml^{-1} ) is

A

6.02×10236.02 \times 10^{23}

B

6.02×10256.02 \times 10^{25}

C

6.02×10246.02 \times 10^{24}

D

6.02�10236.02 � 10^{23}

Answer

6.02×10246.02 \times 10^{24}

Explanation

Solution

In 18 mL18\, mL, number of moles of H2O= mass of H2O molecular mass H_{2} O=\frac{\text { mass of } H_{2} O}{\text { molecular mass }} = density × velume  molecular mass =\frac{\text { density } \times \text { velume }}{\text { molecular mass }} =1×1818=1mol=\frac{1 \times 18}{18}=1 mol ∵\because Number of moles of H2OH _{2} O in 1 mol1\, mol =6.022×1023=6.022 \times 10^{23} and number of e−e^{-} in 11 molecule of H2OH _{2} O =1×2+8=10=1 \times 2+8=10 ∴\therefore Number of e−e^{-} in 1 mole1\, mole of H2OH _{2} O =6.022×1023×10=6.022 \times 10^{23} \times 10 =6.022×1024=6.022 \times 10^{24}