A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C.

Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.


The molar concentration of the gene fragment

 straight M subscript straight B straight space equals straight space fraction numerator straight W subscript straight B cross times straight space straight R cross times straight T over denominator straight pi straight space cross times straight space straight V end fraction

Here, WB = 8.95 mg = 8.95 x 10-3 g, R = 0.0821 L atm Mol-1 K-1

T = 250C = (25 + 273) K = 298 K. = 0.355torr =0.355/760 atm

V = 35 mL = 35 x 10-3 L

Substituting these values in the above equation, we get

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