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Math  /  Algebra

Question0.302 grams of an antibiotic was dissolved in enough water at 23.6∘C23.6^{\circ} \mathrm{C} to make 500.0 mL of solution. The solution has an osmotic pressure of 8.34 mm Hg . What is the molar mass of the antibiotic? Show your work.

Studdy Solution

STEP 1

1. The osmotic pressure of a solution can be calculated using the formula Π=MRT\Pi = MRT, where Π\Pi is the osmotic pressure, MM is the molarity, RR is the gas constant, and TT is the temperature in Kelvin.
2. The molarity MM is given by the ratio of the number of moles of solute to the volume of the solution in liters.
3. The number of moles of solute is the mass of the solute divided by its molar mass.
4. The gas constant RR is 0.08210.0821 L·atm/(mol·K).
5. The temperature must be converted from degrees Celsius to Kelvin by adding 273.15273.15.

STEP 2

1. Convert the osmotic pressure from mm Hg to atm.
2. Convert the temperature from degrees Celsius to Kelvin.
3. Use the osmotic pressure formula to solve for the molarity MM.
4. Calculate the number of moles of the antibiotic.
5. Determine the molar mass of the antibiotic using the mass and the number of moles.

STEP 3

Convert the osmotic pressure from mm Hg to atm. The conversion factor is 1 atm=760 mm Hg1 \text{ atm} = 760 \text{ mm Hg}.
Π=8.34 mm Hg760 mm Hg/atm=0.010974 atm \Pi = \frac{8.34 \text{ mm Hg}}{760 \text{ mm Hg/atm}} = 0.010974 \text{ atm}

STEP 4

Convert the temperature from degrees Celsius to Kelvin.
T=23.6∘C+273.15=296.75 K T = 23.6^{\circ} \text{C} + 273.15 = 296.75 \text{ K}

STEP 5

Use the osmotic pressure formula Π=MRT\Pi = MRT to solve for the molarity MM.
0.010974=M⋅0.0821⋅296.75 0.010974 = M \cdot 0.0821 \cdot 296.75
M=0.0109740.0821⋅296.75 M = \frac{0.010974}{0.0821 \cdot 296.75}
M≈0.000452 mol/L M \approx 0.000452 \text{ mol/L}

STEP 6

Calculate the number of moles of the antibiotic. The volume of the solution is 500.0500.0 mL, which is 0.5000.500 L.
Number of moles=M×Volume \text{Number of moles} = M \times \text{Volume}
Number of moles=0.000452 mol/L×0.500 L \text{Number of moles} = 0.000452 \text{ mol/L} \times 0.500 \text{ L}
Number of moles≈0.000226 mol \text{Number of moles} \approx 0.000226 \text{ mol}

STEP 7

Determine the molar mass of the antibiotic using the mass and the number of moles.
Molar mass=Mass of soluteNumber of moles \text{Molar mass} = \frac{\text{Mass of solute}}{\text{Number of moles}}
Molar mass=0.302 g0.000226 mol \text{Molar mass} = \frac{0.302 \text{ g}}{0.000226 \text{ mol}}
Molar mass≈1336.28 g/mol \text{Molar mass} \approx 1336.28 \text{ g/mol}
The molar mass of the antibiotic is approximately 1336.281336.28 g/mol.

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