By Boyles Law (P1V1=P2V2), substituting values in and solving for V2, we find that the new volume is 3.6 L<span />
(a) One form of the Clausius-Clapeyron equation is
ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:
Solving for ΔHv:
- ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
- ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:
- ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
- 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
- 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.
M1v1=m2v2
m2=(m1v1)/v2
Where m is the molarities and v is the volumes
<span>m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
m2=0.234 M
so the answer is C: 0.234 M</span>
Answer:
The molar mass of Mg(NO₃)₂, 148.3 g/mol.
Explanation:
Step 1: Given data
- Mass of Mg(NO₃)₂ (solute): 42.0 g
- Volume of solution: 259 mL = 0.259 L
Step 2: Calculate the moles of solute
To calculate the moles of solute, we need to know the molar mass of Mg(NO₃)₂, 148.3 g/mol.
42.0 g × 1 mol/148.3 g = 0.283 mol
Step 3: Calculate the molarity of the solution
M = moles of solute / liters of solution
M = 0.283 mol / 0.259 L
M = 1.09 M