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A solution prepared by dissolving 8.95 mg of a gene fragment in35.0 ml of water has an osmotic pressure of 0.335 ton at 25^(@)C. Assuming thegene fragment is a non-electrolyse, determine the molar mass. |
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Answer» Solution :Mass of gene fragment = 8.95 mg = `8.95 xx 10^(-3) g` Volume of water = `35.0 ML = 35 xx 10^(-3)` L `pi = 0.335` ton = 0.335/760 atm Temp = 25 + 273 = 298 K `pi = (W_(B) RT)/(M_(B) xx V)` `(0.335)/(760) = (8.95 xx 10^(-3) xx 0.0821 xx 298)/(M_(B) xx 35 xx 10^(-3))` `M_(B) = 141933 g MOL^(-3)` |
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