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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

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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 =

1×10⁻⁷ kg will have =

number of viruses in 1 mole = 6.022×10²³ viruses

number of viruses in mole =

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

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 =

1×10⁻⁷ kg will have =

number of viruses in 1 mole = 6.022×10²³ viruses

number of viruses in mole =

So number of viruses present in 1 cc will be