Specific volume of cylinderical virus particle is6.02x10 cc/g, whose r

1.	Specific volume of cylinderical virus particle is6.02x10 cc/g, whose radius and length are7A and 10A respectively. If NA -6.02x102318.find molecualr weight of virus. IAIPMT 20011(A) 15.4 kg/mol (B) 1.54x104 kg/moi(C) 3.08x104 kg/moł (D) 3.08x10 kg/mol
Answer» Volume of one virus = voume of cylinder = π r²h = 22/7 × 7 × 7 × 10 A°³ =22 × 7 × 10 A°³ =1540 × 10^-30 m³ =1.54 × 10^-27 m³ so, volume of one mole of virus = 6.023 × 10²³ × volume of one virus = 6.023 × 10²³ × 1.54 × 10^-27 m³ = 9.27542 × 10^-4 m³ = 9.27542 × 10² cm³ now, molecular weight = volume of 1mole/ specific volume = 9.27542 ×10² / 6.02 × 10^-2 g/mol= 1.54 × 10⁴ g/mol

Specific volume of cylinderical virus particle is6.02x10 cc/g, whose radius and length are7A and 10A respectively. If NA -6.02x102318.find molecualr weight of virus. IAIPMT 20011(A) 15.4 kg/mol (B) 1.54x104 kg/moi(C) 3.08x104 kg/moł (D) 3.08x10 kg/mol

Answer»

Volume of one virus = voume of cylinder = π r²h

= 22/7 × 7 × 7 × 10 A°³

=22 × 7 × 10 A°³

=1540 × 10^-30 m³

=1.54 × 10^-27 m³

so, volume of one mole of virus = 6.023 × 10²³ × volume of one virus

= 6.023 × 10²³ × 1.54 × 10^-27 m³

= 9.27542 × 10^-4 m³

= 9.27542 × 10² cm³

now, molecular weight = volume of 1mole/ specific volume

= 9.27542 ×10² / 6.02 × 10^-2 g/mol= 1.54 × 10⁴ g/mol