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 ton at `25^(@)C`. Assuming the gene fragment is a non-electrolyse, determine the molar mass.

Explanation

Mass of gene fragment = 8.95 mg
= `8.95 xx 10^(-3) g`
Volume of water = `35.0 ml = 35 xx 10^(-3)` L
`pi = 0.335` ton = 0.335/760 atm
Temp = 25 + 273 = 298 K
`pi = (W_(B) RT)/(M_(B) xx V)`
`(0.335)/(760) = (8.95 xx 10^(-3) xx 0.0821 xx 298)/(M_(B) xx 35 xx 10^(-3))`
`M_(B) = 141933 g mol^(-3)`