The density of concentrated ammonia, which is 28.0% w/w NH3, is 0.899 g/mL. What volume of this reagent should be diluted to 1.0 × 103 mL to make a solution that is 0.036 M in NH3?
The correct answer and explanation is :
To find the volume of 28.0% (w/w) concentrated ammonia needed to prepare 1000 mL (1.0 × 10³ mL) of a 0.036 M ammonia solution, we need to go step-by-step:
Step 1: Determine the molarity of the concentrated ammonia
We’re given:
- 28.0% w/w NH₃: This means 28 g of NH₃ per 100 g of solution.
- Density = 0.899 g/mL: So 1 mL of solution weighs 0.899 g.
Let’s find out how many moles of NH₃ are present in 1 liter (1000 mL) of the concentrated solution.
Mass of 1 mL of solution = 0.899 g
Mass of 1000 mL = 0.899 g/mL × 1000 mL = 899 g
Now, calculate the mass of NH₃ in this 899 g of solution:
[
\text{Mass of NH₃} = 28\% \times 899\ \text{g} = 0.28 \times 899 = 251.72\ \text{g NH₃}
]
Convert this to moles using the molar mass of NH₃ (17.03 g/mol):
[
\text{Moles of NH₃} = \frac{251.72\ \text{g}}{17.03\ \text{g/mol}} \approx 14.78\ \text{mol}
]
So, molarity of concentrated NH₃ is:
[
\frac{14.78\ \text{mol}}{1\ \text{L}} = 14.78\ \text{M}
]
Step 2: Use dilution formula to find required volume
[
M_1 V_1 = M_2 V_2
]
Where:
- (M_1 = 14.78\ \text{M}) (concentrated NH₃)
- (V_1 =) unknown (volume to dilute)
- (M_2 = 0.036\ \text{M}) (final concentration)
- (V_2 = 1000\ \text{mL}) (final volume)
[
14.78 \times V_1 = 0.036 \times 1000
]
[
V_1 = \frac{36}{14.78} \approx 2.43\ \text{mL}
]
✅ Final Answer: 2.43 mL of concentrated ammonia solution
Explanation
This problem involves using percentage concentration, density, and molarity concepts to perform a dilution. First, we used the w/w % and density to determine how many grams and moles of ammonia are present in 1 liter of the concentrated solution. This gave us the molarity of the concentrated reagent. Then, we used the dilution formula (M_1V_1 = M_2V_2) to find how much of the concentrated solution was required to prepare the target dilute solution. The final step involved simple algebra to solve for the unknown volume. This is a classic application of solution chemistry in lab preparations.