A 0.15 -g sample of a purified protein is dissolved in water to give 2.0 \(\mathrm{mL}\) of solution. The osmotic pressure is found to be 18.6 torr at \(25^{\circ} \mathrm{C}\) . Calculate the protein's molar mass.

Short Answer

Expert verified
The molar mass of the protein is approximately 372 g/mol.

Step by step solution

01

Convert the given data to appropriate units

We need to make sure all the given data are in the correct units. The osmotic pressure is given in torr, which we will need to convert to atm, and the temperature is given in Celsius, which we need to convert to Kelvin. Osmotic pressure in atm: \(18.6 \frac{torr}{1} \times \frac{1 atm}{760 torr} = 0.0245 atm\) Temperature in Kelvin: \(25^{\circ} C + 273.15 = 298.15 K\) Volume in liters: \(2.0\, mL = 0.002\, L\)
02

Rearrange the osmotic pressure equation to find moles

We need to find the number of moles of the protein in the solution. Rearranging the formula for osmotic pressure, we get: \(n = \frac{πV}{RT}\) Use the converted data from step 1 and solve for the number of moles: \(n = \frac{0.0245\, atm \times 0.002\, L}{(0.0821\, L\, atm/mol\, K)(298.15\, K)} = 0.000403\, mol\)
03

Calculate the molar mass of the protein

The molar mass of the protein can be calculated by dividing its mass by the number of moles: Molar mass = \(\frac{mass}{moles}\) Molar mass = \(\frac{0.15\, g}{0.000403\, mol} = 372\, g/mol\) The molar mass of the protein is approximately 372 g/mol.

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