Answer to Question #6249 in Organic Chemistry for Ina
A solution of 30 percent ethanol (C2 H3 OH) by weight has a density of 0.96 g/mL at 25 percent C. Find its molality and molarity.
The mass fraction wi is the fraction of one substance with mass mi to the mass of the total mixture mtot, defined as: wi=mi/mtot. The mass of the ethanol-water mixture is mmix=d*V. (d=mixture density, V=mixture volume). Therefore, wi=mi/d*V. Hence, V=mi/d*wi. mi=ni*Mi, so V=ni*Mi/d*wi(in millilitres). The molar concentration, ci is defined as the amount of a constituent ni divided by the volume of the mixture V(in Litres): ci=ni/V And 1 millilitre is 0.001 litres. Putting it all together we have: c=n*d*w/n*M=d*w/1000*M Ethanol have a chemical formula C2H5OH. Its molar mass is 46 g/mol. The solution molarity is: c=0.96g/mL*0.3/0.001*46g/mol=6.26mol/litre=6.26M
The molality, b, of a solvent/solute combination is defined as the amount of solute, nsolute, divided by the mass of the solvent, msolvent (not the mass of the solution): b= nsolute/ msolvent. The amount of solute is nsolute=mi/Mi. mi=wi*mtot, and mtot=mi+msolvent, so msolvent=mtot-mi=(1-w)mtot. Putting it all together we have: b=ni/msolv=mi/Mmsolv=w*mtot/M*(1-w)*mtot=w/M*(1-w) The solution molality is: b=0.3/46g/mol*(1-0.3)=0.3/46*0.7=0.0093 Answer: The molarity is 6.26M, the molality is 0.0093.