You require a solution containing 1.15 mM of sodium hydroxide. What is the maximum volume of solution (L) that can be prepared from 0.27 kg of sodium hydroxide (whole number)?
The correct answer and explanation is:
✅ Correct Answer: 5,870 L
Explanation:
To determine the maximum volume of a 1.15 mM sodium hydroxide (NaOH) solution that can be prepared from 0.27 kg of NaOH, follow these steps:
Step 1: Convert mass to moles
Given:
- Mass of NaOH = 0.27 kg = 270 g
- Molar mass of NaOH = 22.99 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g/mol ≈ 40 g/mol
Moles of NaOH=MassMolar mass=270 g40 g/mol=6.75 mol\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{270\text{ g}}{40\text{ g/mol}} = 6.75\text{ mol}
Step 2: Use molarity formula
Molarity (M) = moles / volume (L)
We rearrange to find volume: Volume (L)=MolesMolarity=6.75 mol1.15×10−3 mol/L=6.750.00115≈5869.565 L\text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{6.75\ \text{mol}}{1.15 \times 10^{-3}\ \text{mol/L}} = \frac{6.75}{0.00115} \approx 5869.565 \text{ L}
Step 3: Round to whole number
Since the question asks for the maximum volume in whole liters: 5,870 L\boxed{5,870\ \text{L}}
Conclusion:
From 0.27 kg of sodium hydroxide, you can prepare a maximum of 5,870 liters of a 1.15 mM NaOH solution. This calculation is vital in chemical solution preparation where concentration accuracy and resource efficiency are critical in labs, industries, and education.