To solve this problem, we can use the concept of dilution, which is described by the equation:

$$ C_1 V_1 = C_2 V_2 $$

where:
- $ C_1 $ is the concentration of the stock solution (1.63 M),
- $ V_1 $ is the volume of the stock solution needed,
- $ C_2 $ is the concentration of the diluted solution (0.155 M),
- $ V_2 $ is the volume of the diluted solution (2.00 L).

We need to find $ V_1 $. Rearrange the equation to solve for $ V_1 $:

$$ V_1 = \frac{C_2 V_2}{C_1} $$

Substitute the given values into the equation:

$$ V_1 = \frac{0.155 \, \text{M} \times 2.00 \, \text{L}}{1.63 \, \text{M}} $$

Calculate the value:

$$ V_1 = \frac{0.310 \, \text{M} \cdot \text{L}}{1.63 \, \text{M}} $$

$$ V_1 \approx 0.190 \, \text{L} $$

Therefore, the volume of the 1.63 M stock solution required to prepare 2.00 L of the 0.155 M saline solution is approximately 0.190 liters.

Question

To solve this problem, we can use the concept of dilution, which is described by the equation:

$$ C_1 V_1 = C_2 V_2 $$

where:
- $ C_1 $ is the concentration of the stock solution (1.63 M),
- $ V_1 $ is the volume of the stock solution needed,
- $ C_2 $ is the concentration of the diluted solution (0.155 M),
- $ V_2 $ is the volume of the diluted solution (2.00 L).

We need to find $ V_1 $. Rearrange the equation to solve for $ V_1 $:

$$ V_1 = \frac{C_2 V_2}{C_1} $$

Substitute the given values into the equation:

$$ V_1 = \frac{0.155 \, \text{M} \times 2.00 \, \text{L}}{1.63 \, \text{M}} $$

Calculate the value:

$$ V_1 = \frac{0.310 \, \text{M} \cdot \text{L}}{1.63 \, \text{M}} $$

$$ V_1 \approx 0.190 \, \text{L} $$

Therefore, the volume of the 1.63 M stock solution required to prepare 2.00 L of the 0.155 M saline solution is approximately 0.190 liters.

Knowee AI · Accepted Answer