Q19 of 50 Page 1

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.

To calculate osmotic pressure, we know that,



[As, n = ]


Here, π is the osmotic pressure


n = Number of moles of solute


w = Weight of solute


M = Molecular mass of solute


R = Gas constant = 0.0821 L atm K–1mol–1


t = Temperature



Given,


V = 35ml. = L


T = 298K.


π = 0.335torr. = atm.


w = 8.95×10-3gm.


So,


M = (8.95×10-3 × 0.0821 × 298 ×760 × 1000) / (0.335 × 35)


∴ M = = 1.42×104 gm.


So, the molar mass is 1.42×104 gm.


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