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Determine the amount of CaCl_(2)(i=2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at27^(@)C. |
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Answer» Solution :`pi=iCRT=i(n)/(V)RT"or"n=(pixxV)/(ixxRxxT)=("0.75 ATM"xx"2.5 L")/(2.47xx"0.0821 L atm K"^(-1)"mol"^(-1)xx"300 K")="0.0308 mole"` `"Molar mass of "CaCl_(2)=40+2xx35.5="111 g mol"^(-1) therefore"Amount DISSOLVED "=0.0308xx"111 g = 3.42 g."` |
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