61 800
Assignments Done
98,1%
Successfully Done
In May 2018

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.
Expert's answer
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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

Submit
Privacy policy Terms and Conditions