Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>
Answer:
398 mL
Explanation:
Using the equation for molarity,
C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L
V₂ = V₁ + V' where V' = volume of water added.
So, From C₁V₁ = C₂V₂
V₁ = C₂V₂/C₁
= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L
= 0.875 mol/8.61 mol/L
= 0.102 L
So, V₂ = V₁ + V'
0.5 L = 0.102 L + V'
V' = 0.5 L - 0.102 L
= 0.398 L
= 398 mL
So, we need to add 398 mL of water to the nitric solution.
