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.

Explanation

We are given that
Mass of gene fragment dissolved (W) = 8.95 mg = `8/.95xx10^(-3)g`
Volume of solution (V) = 35 ml = `35xx10^(-3)L`
Osmotic pressure `(pi)` = 0.335 torr
We are to find out Molar mass (M) of the gene fragment we know that
`pi=CRT=n/(V)RT=w/(pi)xx1/(V)RT` `(becausen=W/(M))`
`becauseM=W/(pi)xx1/(V)RT=(8.95xx10^(-3))/(0.335)xx1/(35xx10^(-3))xx0.0821xx298K=18.72g`