A 20 ml solution of sugar with a concentration of 1.0M is diluted to 120 mL, what is the new
concentration?

M₂  = 0.16 M

Explanation:

Given data:

Initial volume of solution = 20 mL

Initial concentration = 1.0 M

Final volume = 120 mL

Final/new concentration = ?

Solution:

Formula:

M₁V₁     =    M₂V₂

by putting values,

1.0 M × 20 mL = M₂ × 120 mL

M₂  = 1.0 M × 20 mL / 120 mL

M₂  = 20 M.mL /120 mL

M₂  = 0.16 M

[PCl₃] = [Cl₂] = 0.035M

[PCl₅] = 0.059M

Explanation:

Based on the chemical reaction, Kc is:

Kc = 49 = [PCl₅] / [PCl₃] [Cl₂]

Where [] are the concentrations in equilibrium of each species

The initial concentration of the reactants is:

[PCl₃] = [Cl₂] = 0.75mol / 8.0L =  0.094M

The reactants must be consumed whereas PCl₅ will be produced. That is:

[PCl₃] = [Cl₂] = 0.094M - X

[PCl₅] = X

Where X is the reaction coordinate

Replacing in Kc expression:

49 = [X] / [0.094 - X] [0.094 - X]

49 = [X] / 0.008836 - 0.188 X + X²

0.432964 - 9.212 X + 49 X² = X

0.432964 - 10.212 X + 49 X² = 0

Solving for X

X = 0.059M → Right solution

X = 0.15M → False solution. Will produce negative concentrations

Replacing:

[PCl₃] = [Cl₂] = 0.094M - 0.059M = 0.035M

[PCl₅] = 0.059M