to what temperature (C) would a solution containing C6H12O6 in 200g of water have to be heated to have a vp of 89.5 mmHg?
To find the temperature at which a solution containing C6H12O6 (glucose) in 200 grams of water would have a vapor pressure (vp) of 89.5 mmHg, you will need to use the Clausius-Clapeyron equation. This equation relates temperature, vapor pressure, and enthalpy of vaporization.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures respectively,
ΔHvap is the enthalpy of vaporization,
R is the ideal gas constant (8.314 J/mol·K),
T1 and T2 are the initial and final temperatures respectively.
Since we want to find the temperature (T2) at which the vapor pressure (P2) is 89.5 mmHg in water, we can rearrange the equation and solve for T2:
ln(89.5/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Now, we need to gather the necessary information to plug into the equation.
1. Find the initial vapor pressure (P1) of water at the initial temperature (T1). Since the solution contains only glucose, we can approximate the vapor pressure of water as the vapor pressure of pure water at this temperature. You can look up the vapor pressure of water in a table or use an online resource.
2. Determine the enthalpy of vaporization (ΔHvap) of water. This value represents the energy required to convert one mole of liquid water to gas at a specific temperature.
3. Substitute the values into the equation and solve for T2:
ln(89.5/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Solving for T2 will give you the temperature at which the solution containing glucose and water would have a vapor pressure of 89.5 mmHg.