The volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
<h3>How to calculate volume?</h3>
The concentration of a solution can be calculated using the following formula:
C1V1 = C2V2
Where;
- C1 = initial concentration = 1.5M
- C2 = final concentration = 0.54M
- V1 = initial volume = ?
- V2 = final volume = 0.18L
1.5 × V1 = 0.54 × 0.18
1.5V1 = 0.0972
V1 = 0.0972 ÷ 1.5
V1 = 0.065L
Therefore, the volume of the stock solution that has a concentration of 1.5 M SO2 and is diluted to a 0.54 M solution with a volume of 0.18 L is 0.065L.
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Answer:
[NH₃] → 3.24 M
Explanation:
Our solute: Ammonia
Our solvent: Water
Solution's mass = Mass of solute + Mass of solvent
Solution's mass = 15 g + 250 g = 265g
We use density to determine, the volume.
D = mass /volume → Volume = m / D → 265 g /0.974 g/mL = 272.07 mL.
We convert the mL to L → 272.07 mL . 1L /1000mL = 0.27207 L
To determine molarity we need the moles of solute in 1 L of solution.
Moles of solute are: 15g / 17g/mol = 0.882 moles
[NH₃] = 0.882mol /0.27207 L → 3.24 M
Number of moles = volume / (molar volume)
Molar volume at stp = 22.4 dm^3
Volume = no of moles × molar volume
= 0.987 × 22.4
= 22.1088 dm^3
= 22108.8 cm^3
Hope it helped!
Answer:
because they need only one electron
