Question

The no. of molecules present in one drop (0.05ml) of benzene (density=0.78g/ml) is nearly equal to (${N_A} = 6 \times {10^{23}}$ ):
A) $6 \times {10^{23}}$
B) $3 \times {10^{21}}$
C) $3 \times {10^{20}}$
D) $6 \times {10^{19}}$

Answer 1

Hint: First let us find the mass of one drop of Benzene and then find the number of moles of Benzene in a given drop. And then find the number of molecules.

Complete answer:
We are given that there is one drop of benzene
And density of benzene=0.78g/ml
And volume of one drop = 0.05ml
Therefore, we are going to find mass of one drop of benzene using the formula
Mass of 1 drop of benzene =volume of one drop$ \times $ density of benzene
Now by putting required values we get,
 Mass of one drop of benzene= $0.05ml \times $0.78g ml
Mass of one drop of benzene=0.039g
We know that,
No. of Moles of benzene = Mass of one drop of benzene$ \div $Molecular mass of benzene
And here molecular mass of benzene is also required which we will find using the given formula,
 Molecular mass of benzene=$6 \times $Atomic mass of carbon+$6 \times $Atomic mass of hydrogen
(As per the molecular formula of benzene that is ${C_6}{H_6}$ )
Therefore molecular mass of benzene by putting values in formula=6$ \times $12+6$ \times $1
And Molecular Mass of benzene =78g/mole
Therefore, by putting the calculated values in above formula we get,
Moles of benzene=$0.039g \div $78g/mole
Moles of benzene= $5 \times {10^{ - 4}}$
And as 1 mole of benzene contains ${N_A} = 6 \times {10^{23}}$molecules
Therefore No. of molecules in $5 \times {10^{ - 4}}$ no. of moles of benzene or we can say no. of molecules present in one drop of benzene=no. of molecules in 1 mole of benzene $ \times $no. moles of benzene in one drop
 =$5 \times {10^{ - 4}}$$ \times $$6 \times {10^{23}}$ molecules
 =$3 \times {10^{20}}$molecules
Therefore, the no. of molecules present in one drop (0.05ml) of benzene (density=0.78g/ml) is equal to$3 \times {10^{20}}$

Therefore, the correct answer is option ‘C’.

Note:Here we have to remember that the No. of molecules = no. of molecules of that compound in one mole$ \times $no. of moles of in one drop.