Answer to Question #6249 in Organic Chemistry for Ina

Question #6249
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.
1
Expert's answer
2012-02-07T14:00:26-0500
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.

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