Answer to Question #283249 in Inorganic Chemistry for Loco Kilo

Question

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.

 

Answer 1

$\Delta\>T_b=k_b.m.i$

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

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

(For water $=0.515°C/m)$

$m=$ molarity of the solute

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

$0.34=0.515×m×1$

$m=0.660194\>moles$

Gram of Glucose in 1kg of water

$\frac{1000}{150}×18=120g$

$0.660194\>moles\equiv120g$

$1\>mole\equiv$

$\frac{120×1}{0.660194}=181.76g$