Question

In a given species of tobbaco there is 0.1 mg of virus per c.c. The mass of virus is $4\times {10}^7$ Kg per kilomol. The number of the molecules of virus present in 1 c.c. will be

• A. ${10}^{13}$
• B. ${10}^{11}$
• C. $1.5\times {10}^{12}$
• D. ${10}^{14}$

Answer 1

The correct option is C $1.5\times {10}^{12}$

$m_v=\frac{M}{N_A}=\frac{4\times {10}^7}{6\times {10}^{23}\times {10}^3}massofviruspercc=0.1\times {10}^{-6}kg$

Number of molecules in cc

$n=\frac{m_v}{N_A}=\frac{0.1\times {10}^{-6}\times 6\times {10}^{23}\times {10}^3}{4\times {10}^7}n=1.5\times {10}^{12}$