Question

How much volume of 8M Hcl is required to prepare 1L 4M Hcl solution?

Answer 1

0.5 L

Answer 2

The principle of dilution states that the number of moles of solute remains constant during the dilution process. The number of moles ($n$) is calculated as the product of molarity ($M$) and volume ($V$), i.e., $n = M \times V$.

Let $V_1$ be the volume of the 8M HCl solution required. The number of moles of HCl in the initial concentrated solution is: $n_1 = M_1 \times V_1 = 8 \text{ M} \times V_1$

The final solution is 1 L of 4M HCl. The number of moles of HCl in the final diluted solution is: $n_2 = M_2 \times V_2 = 4 \text{ M} \times 1 \text{ L}$

Since the number of moles of HCl remains constant during dilution ($n_1 = n_2$): $8 \text{ M} \times V_1 = 4 \text{ M} \times 1 \text{ L}$

Solving for $V_1$: $V_1 = \frac{4 \text{ M} \times 1 \text{ L}}{8 \text{ M}}$ $V_1 = 0.5 \text{ L}$

To express the volume in milliliters, we convert liters to milliliters: $V_1 = 0.5 \text{ L} \times 1000 \text{ mL/L} = 500 \text{ mL}$

Therefore, 500 mL of 8M HCl is required.