Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
Answer:
13.20
Explanation:
Step 1: Calculate the moles of Ba(OH)₂
The molar mass of Ba(OH)₂ is 171.34 g/mol.
0.797 g × 1 mol/171.34 g = 4.65 × 10⁻³ mol
Step 2: Calculate the molar concentration of Ba(OH)₂
Molarity is equal to the moles of solute divided by the liters of solution.
[Ba(OH)₂] = 4.65 × 10⁻³ mol/60 × 10⁻³ L = 0.078 M
Step 3: Calculate [OH⁻]
Ba(OH)₂ is a strong base according to the following equation.
Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻
The concentration of OH⁻ is 2/1 × 0.078 M = 0.16 M
Step 4: Calculate the pOH
pOH = -log OH⁻ = -log 0.16 = 0.80
Step 5: Calculate the pH
We will use the following expression.
pH + pOH = 14
pH = 14 - 0.80 = 13.20
P = 1.5atm ≈ 1519.88hPa
V = 8.56L
R = 83.1 [hPa*L] / [mol*K]
T = 0°C =273K
pV = nRT |:RT
n = pV / RT
n = [1519.88hPa*8.56L] / [83.1 [hPa*L] / <span>[mol*K] * 273K]
n </span>≈ <u>0.57mol</u><span><u> </u></span>
