Question

A student placed 19.0 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.-mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 25.0-mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution

Answer 1

Concentration of glucose slotion = (19/180)*(1000/100)

                                                            = 1.0555 mol/L

M1 V1 = M2 V2

1.0555 * 25 = M2 * 500

M2 = 0.0528 mol/L

0.0528 mole in 1 liter of solution then

0.00528 mole of glucose present in 100 mL solution.

mass of glucose = 0.00528 * 180 = 0.9504 g