Question #283249

A solution was prepared by dissolving 18.00g of glucose in 150.0g of water. The resulting solution was found to have a boiling point of 0.34°C higher than its pure solvent boiling point at 1 atm. Calculate the molar mass of glucose. Glucose is a molecular solid that is present in individual molecules in solution.

 



1
Expert's answer
2021-12-29T12:26:02-0500



ΔTb=kb.m.i\Delta\>T_b=k_b.m.i


Where ΔTb=\Delta\>T_b= boiling point elevation

kb=k_b= boiling point elevation constant which depend on solvent


(For water =0.515°C/m)=0.515°C/m)

m=m= molarity of the solute

i=i= number of particles formed when compound dissolves.



0.34=0.515×m×10.34=0.515×m×1

m=0.660194molesm=0.660194\>moles


Gram of Glucose in 1kg of water


1000150×18=120g\frac{1000}{150}×18=120g


0.660194moles120g0.660194\>moles\equiv120g

1mole1\>mole\equiv



120×10.660194=181.76g\frac{120×1}{0.660194}=181.76g


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 the first!
LATEST TUTORIALS
APPROVED BY CLIENTS