An analytical chemist evaluates four aqueous glucose (C H O ) solutions—labeled W, X, Y
CHEMISTRY · Multi-part · AP · Chemistry · Atomic Structure and Properties · Atomic Structure and Properties
Problem
An analytical chemist evaluates four aqueous glucose (C6H12O6) solutions—labeled W, X, Y, and Z—by evaporating exactly 10.00 mL of each sample to dryness and weighing the residue. The masses of the solid residue are recorded in the table.
Sample | Mass of residue (g)
W | 0.135
X | 0.210
Y | 0.112
Z | 0.175
Assume all residue is pure glucose. (Molar mass of glucose = 180 g mol−1.)
(a) Rank the four samples from highest to lowest glucose concentration, and justify your ordering qualitatively before performing any calculations.
(b) Derive an algebraic expression that relates the molarity M of a glucose solution to the measured mass m (g) of residue obtained from evaporating a fixed sample volume V (L). Clearly state any assumptions.
(c) Using your expression from part (b), calculate the molarity of sample X.
(d) If a small amount of water remained in the evaporating dish for sample Y, explain qualitatively how this procedural error would affect the calculated molarity for that sample and the overall ranking from part (a).