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.

Solution

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, W_B = 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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Solutions

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.

Some More Questions From Solutions Chapter

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

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NCERT Sample Papers

Entrance Exams Preparation

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Solutions

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.

Some More Questions From Solutions Chapter

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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.