A Solution Contains 10.0% (By Mass) Or (Mass/Mass) NaBr (Sodium Bromide) (Molar Mass= 102 89 G/Mol). The Density Of The Solution Is 1 046 G/Cm What Is The Molarity Of The NaBr Solution?
A 0.610
B. 0.970
C 0.280
D. 1.02 M
E. 0.583
The correct answer and explanation is :
To find the molarity (M) of the NaBr solution, follow these steps:
Step 1: Understand Given Data
- Mass percent of NaBr = 10.0% (which means 10.0 g of NaBr in 100 g of solution)
- Molar mass of NaBr = 102.89 g/mol
- Density of the solution = 1.046 g/cm³ (which is equivalent to 1.046 g/mL)
Step 2: Determine the Volume of the Solution
We assume 100 g of solution. Using density:
[
\text{Volume} = \frac{\text{Mass of solution}}{\text{Density}}
]
[
\text{Volume} = \frac{100 \text{ g}}{1.046 \text{ g/mL}} = 95.60 \text{ mL} = 0.0956 \text{ L}
]
Step 3: Find the Moles of NaBr
Since 10.0 g of NaBr is present:
[
\text{Moles of NaBr} = \frac{\text{Mass of NaBr}}{\text{Molar mass of NaBr}}
]
[
\text{Moles of NaBr} = \frac{10.0 \text{ g}}{102.89 \text{ g/mol}} = 0.0972 \text{ moles}
]
Step 4: Calculate Molarity
Molarity (M) is defined as:
[
M = \frac{\text{Moles of solute}}{\text{Liters of solution}}
]
[
M = \frac{0.0972}{0.0956} = 1.02 \text{ M}
]
Final Answer:
The correct answer is D. 1.02 M
Explanation (300 words)
Molarity (M) represents the concentration of a solution, measured as moles of solute per liter of solution. In this problem, we were given a 10.0% NaBr solution by mass, meaning that for every 100 g of solution, 10.0 g is NaBr. The density of the solution is 1.046 g/mL, allowing us to determine the volume of 100 g of solution.
To find the molarity, we first calculated the volume of 100 g of solution using the density formula, giving 95.60 mL (or 0.0956 L). Then, we calculated the moles of NaBr using its molar mass (102.89 g/mol), resulting in 0.0972 moles.
Finally, dividing the moles of NaBr (0.0972 moles) by the volume in liters (0.0956 L) gave us the molarity:
[
M = \frac{0.0972}{0.0956} = 1.02 M
]
This means that the solution has 1.02 moles of NaBr per liter, making the correct answer option D (1.02 M).
