A 5.7 x 10-2 mg sample of a protein is dissolved in water to make 0.25 ml of sol

Question

A 5.7 x 10-2 mg sample of a protein is dissolved in water to make 0.25 ml of solution. The osmotic pressure of the solution is 0.56 torr at 25C. What is the molar mass of the protein?

Explanation / Answer

molarity =osmotic pressure / R * T ------- Use R=62.36 L torr K^-1 ml^-1 , T=25+273 = 298 K, osmotic pressure =0.56 torr-------> Now put these values in mentioned equation and solve for molarity ------ molarity = 0.56 / 62.36 * 298 = 1.0 * 10^-7 -------- molarity = mass of solute / (molar mass of solute * volume in L) --------> molar mass of solute = mass of solute in g / (molarity * volume in L)----- mass of solute = 5.7 x 10-2 mg = 5.7 x 10^-5 g and volume = 0.25 mL = 0.00025 L ------Put these values in molar mass formula mentioned above and solve for molar mass -----molar mass of solute = 5.7 x 10^-5 g /( 1.0 * 10^-7 * 0.00025 L) = 2.28 *10^6 g / mol = 2.3 *10^6 g / mol

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