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What is the molality of solution consisting of 33.0 mL of benzene (C_(6)H_(6),d=0.877 g mL^(-1)) in 187 mL of hexane (C_(6)H_(14), d= 0.660 g ml^(-1))? |
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Answer» Mass of benzene `=(28.941)/78 = 0.371` Mass of hexane `=187 xx 0.66 = 123.42`g `m=(0.371 xx 1000)/(123.42) = 3.006` |
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