the volume of ethyl alcohol that has to be added to prepare 100 cc of 0.5 M ethyl alcohol solution water is
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density of ethanol = 1.5 g/cm3 , it means that there are 1.5 gram of ethanol in 1 cm3 of solution.
Number of moles of ethanol needed to obtain 100 ml of 0.5 M ethanol = 0.5 mol1000 ml×100 ml =0.05 mol
Hence, the mass of ethanol needed = 0.05 mol x 46 g mol-1 = 2.3 g
volume of original ethanol require = 2.3g /1.5 g cm-3 = 1.53 cm-3
Number of moles of ethanol needed to obtain 100 ml of 0.5 M ethanol = 0.5 mol1000 ml×100 ml =0.05 mol
Hence, the mass of ethanol needed = 0.05 mol x 46 g mol-1 = 2.3 g
volume of original ethanol require = 2.3g /1.5 g cm-3 = 1.53 cm-3
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