1.

A solution prepared by dissolving 8.95 mg of a given fragment in 35.0 mL of has an osmotic pressure of 0.335 torr at `25^(@)C`. Assuming that the given fragment is non-electolyty. Calculate its molar mass.

Answer» `omega = 8.95 mg = 8.95xx 10^(-4) = 0.0895g`.
`T = 25^(@)C = 25 + 273 = 289K`
`V = 35.0mL = (35)/(1000)L,`
`pi = 0.335 "Torr" = (0.335)/(760)"atm"`
Molar Mass `M = (omegaRT)/(piV)`
`= (8.95 xx 10^(-3) xx 0.0821 xx 298)/((0.335)/(760)xx(35)/(1000))=14193 g//mol^(-1)`


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