Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Im working on science too!! I would help you but the attachment isn't pulling up..
Answer:
27.3 kJ/mol
Explanation:
Step 1: Given data
- Temperature 1 (T₁): 298 K
- Vapor pressure 1 (P₁): P₁
- Temperature 2 (T₂): 318 K
- Vapor pressure 2 (P₂): 2 P₁
Step 2: Calculate the enthalpy of vaporization of this liquid
We will use the Clausius–Clapeyron equation.
ln (P₂/P₁) = -ΔHvap/R × (1/T₂ - 1/T₁)
ln 2 = -ΔHvap/(8.314 J/K.mol) × (1/318 K - 1/298 K)
ΔHvap = 2.73 × 10⁴ J/mol = 27.3 kJ/mol
1 ATP = 101.3 kPa
x ATP = 65.78 kPa
cross-multiply
and you'll get x ATP = 0.64936
So, the answer is 0.65 ATP
