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Question: The no. of molecules present in one drop (0.05ml) of benzene (density=0.78g/ml) is nearly equal to (...

The no. of molecules present in one drop (0.05ml) of benzene (density=0.78g/ml) is nearly equal to (NA=6×1023{N_A} = 6 \times {10^{23}} ):
A) 6×10236 \times {10^{23}}
B) 3×10213 \times {10^{21}}
C) 3×10203 \times {10^{20}}
D) 6×10196 \times {10^{19}}

Explanation

Solution

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×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÷\divMolecular mass of benzene
And here molecular mass of benzene is also required which we will find using the given formula,
Molecular mass of benzene=6×6 \times Atomic mass of carbon+6×6 \times Atomic mass of hydrogen
(As per the molecular formula of benzene that is C6H6{C_6}{H_6} )
Therefore molecular mass of benzene by putting values in formula=6×\times12+6×\times1
And Molecular Mass of benzene =78g/mole
Therefore, by putting the calculated values in above formula we get,
Moles of benzene=0.039g÷0.039g \div 78g/mole
Moles of benzene= 5×1045 \times {10^{ - 4}}
And as 1 mole of benzene contains NA=6×1023{N_A} = 6 \times {10^{23}}molecules
Therefore No. of molecules in 5×1045 \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 ×\timesno. moles of benzene in one drop
=5×1045 \times {10^{ - 4}} ×\times 6×10236 \times {10^{23}} molecules
=3×10203 \times {10^{20}}molecules
Therefore, the no. of molecules present in one drop (0.05ml) of benzene (density=0.78g/ml) is equal to3×10203 \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×\timesno. of moles of in one drop.