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Determine the amount of CaCl_(2)(i=2.47) dissolved in 2.5 litre of waster such that its osmotic pressure is 0.75 atm 27^(@)C at. |
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Answer» SOLUTION :`pi = (n)/(V)RT` `pi=i(W)/(MV)RT` `k=(pi MV)/(iRT)` `pi-0.75` atm V = 2.5 L `i=2.47` `T=(27+273)K` = 300 K Here, `R=0.0821 " atm K"^(-1)MOL^(-1)` `M=1xx40+2xx35.5` `= 111 G mol^(-1)` Therefore, `w=(0.75xx111xx2.5)/(2.47xx0.0821xx300)` = 3.42 gram. |
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