Question

When 92 ml of distilled water (density = 1.0 g/cm³) is mixed with 23 grams of NaCl salt, the density of the resultant brine solution is 1.15 g/cm³. The volume of the brine solution (in ml) is _________.

Hint:

Add the mass of water and salt, then divide the total mass by the given brine density.

Answer 1

Correct Answer: 100

Step 1: Find the mass of water used:
The density of distilled water is given as 1.0 g/cm³, and the volume used is 92 ml, where 1 ml equals 1 cm³. \[ \text{Mass of water} = \text{density} \times \text{volume} = 1.0 \ \text{g/cm}^3 \times 92 \ \text{cm}^3 = 92 \ \text{g} \]
Step 2: Find the total mass of the brine solution:
The brine is formed by mixing the water with 23 grams of NaCl salt. \[ \text{Total mass} = \text{mass of water} + \text{mass of salt} = 92 + 23 = 115 \ \text{g} \]
Step 3: Apply the density formula to find the volume of the brine:
The density of the resultant brine solution is given as 1.15 g/cm³. Using \( \text{density} = \dfrac{\text{mass}}{\text{volume}} \), the volume is \[ V = \frac{\text{mass}}{\text{density}} = \frac{115 \ \text{g}}{1.15 \ \text{g/cm}^3} = 100 \ \text{cm}^3 \]
Step 4: State the result:
Since 1 cm³ equals 1 ml, the volume of the brine solution is 100 ml. Notice that the total mass of 115 g is simply the sum of the water and salt masses, while the increased density of 1.15 g/cm³ compared with 1.0 g/cm³ for pure water causes the final volume to come out less than the raw sum of individual volumes would otherwise suggest, which is physically consistent with dissolving salt into water.
Final Answer:
\[ \boxed{100.0 \ \text{ml}} \]