A solution was prepared by dissolving 100 mg of protein X in
100 ml of water.
Molecular weight of protein X is 15,000 Da;
Avogadro’s number = 6.022 × 1023.
Q.49 The number of molecules present in this solution is:
- (A) 40.15 × 1019
- (B) 6.023 × 1019
- (C) 4.015 × 1019
- (D) 0.08 × 1019
Number of Molecules in Protein Solution
This GATE Biotechnology 2013 question tests mole-to-molecule conversion for proteins in biochemistry. The correct answer is option (C).
Calculation Method
Convert mass to moles, then multiply by Avogadro’s number since molecules are total in the solution, independent of volume. Mass = 100 mg = 0.1 g. Moles = mass / MW = 0.1 / 15,000 = 6.667 × 10⁻⁶ mol. Number of molecules = moles × 6.022 × 10²³ ≈ 4.015 × 10¹⁹.
Formula:
[ N = m/M x N_A ]
where ( m ) = 0.1 g, ( M ) = 15,000 g/mol, ( N_A ) = 6.022 × 10²³ ) mol⁻¹.
Option Analysis
| Option | Value | Explanation |
|---|---|---|
| (A) | 40.15 × 10¹⁹ | Overestimates by factor of 10; likely from using 10 mg instead of 0.1 g. Incorrect mass conversion. |
| (B) | 6.023 × 10¹⁹ | Matches Avogadro’s number directly; ignores mass/MW to find moles. Wrong approach. |
| (C) | 4.015 × 10¹⁹ | Correct: (0.1 / 15,000) × 6.022 × 10²³ = 4.015 × 10¹⁹. Matches calculation. |
| (D) | 0.08 × 10¹⁹ | Too low; possibly moles without multiplying by N_A (6.667 × 10⁻⁶ × 10¹⁹ ≈ 0.067, close variant). Incorrect. |
Exam Relevance
This concept applies to bioprocess calculations, enzyme kinetics, and molecular biology dosing. Practice similar problems for GATE BT using consistent units (g, g/mol).
