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.

Explanation

Mass of gene fragment = 8.95 mg
` = 8.95 xx 10^(-3) g`
Volume of water = 35.0mL ` = 35.0 xx 10^(-3) L`
Osmotic pressure , `pi = = 0.335` torr ` = 0.335//760` atm .
Temperature `= 25^(@)C = 273 + 25 = 298K`
` (0.335)/(760) = (8.95 xx 10^(-3) xx 0.0821 xx 298)/(M_(B) xx 35.0 xx 10^(-3))`
`M_(B) = (8.95 xx 10^(-3) xx 0.0821 xx 298 xx 760) /(0.335 xx 35.0 xx 10^(-3))`
` = 14193.3 g mol^(-1) or `
`1.42 xx 10^(4) g mol^(-1)`