At 29.6 degrees C, pure water has a vapor pressure of 31.1 torr. A solution is prepared by adding 86.7 g of "L" a nonvolatile nonelectrolyte to 390.0 g of water. The vapor pressure of the resulting solution is 27.6 torr. Calculate the molar mass of "L"
Formula is:d/po=nL/nw d=31.1-27.6=3.5 po=31.1 nw=390/18=21.7moles 3.5/31.1=nL/21.7 nL=2.442moles=86.7/M M=35.5g/mol
To calculate the molar mass of "L," we can use the equation:
\( \text{Molar Mass} = \frac{{ \text{Mass of solute (g)}}}{{ \text{Moles of solute}}}} \)
First, we need to calculate the number of moles of "L" using the given information.
1. Calculate the number of moles of water:
\( \text{Moles of water} = \frac{{ \text{Mass of water (g)}}}{{ \text{Molar mass of water (g/mol)}}}} \)
The molar mass of water is approximately 18.015 g/mol.
\( \text{Moles of water} = \frac{{390.0 \, \text{g}}}{{18.015 \, \text{g/mol}}}
2. Calculate the number of moles of "L" by subtracting the moles of water from the total moles of the solution:
\( \text{Moles of "L"} = \text{Total moles of solution} - \text{Moles of water} \)
The total moles of the solution can be calculated by dividing the mass of "L" by its molar mass. However, we need to convert the vapor pressures to their corresponding mole fractions first.
3. Calculate the mole fraction of water:
\( \text{Mole fraction of water} = \frac{{\text{Vapor pressure of water}}}{{\text{Vapor pressure of solution}}} \)
\( \text{Mole fraction of water} = \frac{{27.6 \, \text{torr}}}{{31.1 \, \text{torr}}} \)
4. Calculate the mole fraction of "L":
\( \text{Mole fraction of "L"} = 1 - \text{Mole fraction of water} \)
5. Calculate the total moles of the solution:
\( \text{Total moles of solution} = \frac{{\text{Mass of "L" (g)}}}{{\text{Molar mass of "L" (g/mol)}}} \)
6. Finally, calculate the moles of "L" using the formula from step 2.
7. Substitute the calculated values back into the formula for molar mass to find the molar mass of "L".