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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`. |
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Answer» Correct Answer - `W_(2)=3.42g` `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` |
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