Determine the amount of `CaCl_(2)` (i = 2.47) dissolved in 2.5 L of water sucjh that its osmotic pressure is 0.75 atm at `27^(@)C`.

Answer

`W_(2)=3.42g`

Explanation

`Mw` of `CaCl_(2)=40+2xx35.5=111.0 g mol^(-1)`
`pi=iMRT=ixx(n)/(V)RT`
`(or)`
`n=(pixxV)/(ixxRxxT)=(0.75atmxx2.5L)/(2.47xx0.0821LatmK^(-1)mol^(-1)xx300K)`
`=0.0308mol`
`:.` Weight of `CaCl_(2)` dissolved `=nxxMw`
` =0.0308xx111.0g`
`3.42g`