Question

in a given species of tobacco there is 0.1mg of virus per c.c.the mass of virus is 4x10000000 kg per kilomole.the number of the molecules of virus present in 1 c.c. will be? plz explain with solution

Answer 1

mass of virus = 4×10⁷ kg per kilomole
so mass of 1000 moles virus = 4×10⁷ kg
so mass of 1 mole virus = 4×10⁷/1000 =  4×10⁴ kg

mass of virus in 1cc tobacco, = 0.1mg
0.1mg = 1×10⁻⁷ kg
4×10⁴ kg contains 1 mole of virus
1 kg will have = $\frac{1}{4 \times 10^4}\ moles=2.5 \times 10^{-5}\ moles$
1×10⁻⁷ kg will have = $2.5 \times 10^{-5} \times 1 \times 10^{-7}=2.5 \times 10^{-12}\ moles$

number of viruses in 1 mole = 6.022×10²³ viruses
number of viruses in $2.5 \times 10^{-12}$ mole = $2.5 \times 10^{-12} \times 6.022 \times 10^{23}=1.505 \times 10^{12}\ viruses$

So number of viruses present in 1 cc will be 1.505×10¹².