The concentration of sodium iodide used is 15% in acetone. Calculate the number of moles of sodium acetate used in each test tube. Assume the density of acetone = density of the solution. (Look up the density of acetone). Show your work. Density of Acetone= 0.788 g/ml Molar mass of acetone= 58.08 2ml 15% NaI in acetone (0.788g / ml) * ( 1 mol / 58.08 g) = 0.027 moles
The correct answer and explanation is :
Let’s go step-by-step to calculate the number of moles of sodium iodide (NaI) used in each test tube.
Step 1: Given Data
- Concentration of NaI = 15% (w/v), meaning 15 grams of NaI in 100 mL of solution.
- Density of acetone = 0.788 g/mL
- Volume of solution used in each test tube = 2 mL
- Molar mass of acetone = 58.08 g/mol
Step 2: Calculate the Mass of Solution in 2 mL
Since the density of the solution is the same as acetone, we can calculate the mass of the solution in 2 mL. The mass (in grams) is given by:
[
\text{mass of solution} = \text{density} \times \text{volume}
]
[
\text{mass of solution} = 0.788 \, \text{g/mL} \times 2 \, \text{mL} = 1.576 \, \text{g}
]
So, the mass of the solution in 2 mL is 1.576 grams.
Step 3: Calculate the Mass of Sodium Iodide (NaI) in the Solution
The concentration is 15% (w/v), meaning there are 15 grams of NaI in every 100 mL of solution. To find the mass of NaI in 2 mL of solution, use the following relationship:
[
\text{mass of NaI} = \frac{\text{concentration of NaI}}{100} \times \text{volume of solution}
]
[
\text{mass of NaI} = \frac{15 \, \text{g}}{100 \, \text{mL}} \times 2 \, \text{mL} = 0.3 \, \text{g}
]
Thus, the mass of NaI in 2 mL of solution is 0.3 grams.
Step 4: Calculate the Moles of Sodium Iodide (NaI)
To calculate the number of moles of NaI, we use the formula:
[
\text{moles of NaI} = \frac{\text{mass of NaI}}{\text{molar mass of NaI}}
]
The molar mass of NaI is:
[
\text{Molar mass of NaI} = 22.99 \, \text{g/mol (for Na)} + 126.90 \, \text{g/mol (for I)} = 149.89 \, \text{g/mol}
]
Now, calculate the moles:
[
\text{moles of NaI} = \frac{0.3 \, \text{g}}{149.89 \, \text{g/mol}} = 0.0020 \, \text{mol}
]
Final Answer:
The number of moles of sodium iodide used in each test tube (2 mL of 15% NaI in acetone) is 0.0020 moles.
Explanation:
In this problem, the first key step was calculating the mass of the solution in the test tube using the density and volume. Then, using the given concentration of NaI (15% w/v), we determined the mass of NaI present in the 2 mL of solution. Finally, using the molar mass of NaI, we converted the mass of NaI into moles. This process allowed us to find the number of moles of sodium iodide in the 2 mL of acetone solution.