A 5.09 m solution of the solute water dissolved in the solventethanol is availab

Question

A 5.09 m solution of the solute water dissolved in the solventethanol is available. Calculate the mass ( kg ) of the solutionthat must be taken to obtain 0.6350 kg of water.


Molar Mass (g/mol) H2O 18.015 C2H5OH 46.07 Density ( g/mL ) H2O 0.9982 C2H5OH 0.7893 Name/Formula water H2O ethanol A 5.09 m solution of the solute water dissolved in the solventethanol is available. Calculate the mass ( kg ) of the solutionthat must be taken to obtain 0.6350 kg of water.


Molar Mass (g/mol) H2O 18.015 C2H5OH 46.07 Density ( g/mL ) H2O 0.9982 C2H5OH 0.7893 Name/Formula water H2O ethanol

Explanation / Answer

5.09 m solution means 5.09 moles of water are present in 1000g of ethanol Mass of water = moles * molar mass                        =5.09 * 18.015 g/mol                        =91.69 g We require 0.635 kg or 635 g of water 91.69 g of water are present in 1000 g of ethanol 635 g of water will be present in 1000 g * 635 g /91.69 g                                                 =6925.5 g of ethanol. So mass of solution = 6925.5 + 635 g = 7560.5 g                                                           =7.56 kg

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