1.

`2.5 g` of a substance is present in `200 mL` of solution showing the osmotic pressure of `60 cm Hg` at `15^(@)C`. Calculate the molecular weight of substance.What will be the osmotic pressure if temperature is raised to `25^(@)C`?

Answer» Given that, `W_(2)=2.5 g` ,`V=200/1000 L`
`pi=60/70` atm, `T=288 K`
`:. piV=W_(2)/(Mw_2)RT`
`60/70 xx 200/1000 = 2.5/(Mw_2)xx 0.0821 xx288 rArr Mw_(2)=374.38`
Also, `pi_(1)/pi_(2)=T_(1)/T_(2)`
`60/pi_(2)=288/298 rArr pi_(2) = 62.08 cm`


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