Multiple choice

Specific volume of cylindrical virus particle is $6.02 \times10^{2}$ cc/gm. Whose radius and length are 7 & 10 respectively. Find the molecular weight of the virus. [ If, $N_A$ $= 6.02 \times 10^{23}$]

  1. 1.54 kg/mol
  2. 1.54 $\times10^4$ g/mol
  3. 3.08 $\times 10^4$ kg/mol
  4. 3.08 $\times 10^3$ kg/mol
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B Correct answer
AI explanation

The volume of a single cylindrical virus particle is found using the formula V = pi * r^2 * h, which gives V = 3.14 * 7^2 * 10 = 1538.6 cc. Multiplying this volume by the given specific volume of 6.02 * 10^2 cc/gm yields a mass of 1538.6 / 602 = 2.55 * 10^-21 gm per particle. Multiplying this mass by Avogadro's number, 6.02 * 10^23, results in a molecular weight of 2.55 * 10^-21 * 6.02 * 10^23 = 1.54 * 10^4 g/mol.

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