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PTECH Pharmaceutics-I (General, Physical and Dispensing) QUESTION #2987
Question 1
How much $95\%$ alcohol is needed to prepare $500\text{ ml}$ of $70\%$ alcohol?
  • $368.4\text{ ml}$✔️
  • $350\text{ ml}$
  • $420.5\text{ ml}$
  • $300.0\text{ ml}$
Correct Answer Explanation
Using $C_1V_1 = C_2V_2$: $95 \times V_1 = 70 \times 500 \implies V_1 = 35000 / 95 \approx 368.4\text{ ml}$.