The molarity of an aqueous solution of glucose (C₆H₁₂O₆) is 0.01. To 200 mL of the solution, which of the following should be carried out to make it 0.02 M?
I. Evaporate 50 mL of solution
II. Add 0.180 g of glucose and then evaporate 50 mL of solution
III. Add 50 mL of water
The correct option is:

Since the molarity of glucose has to be increased (from 0.01 M to 0.02 M), so this can be carried out either by evaporating the solution or by adding some more glucose.

So, the only possible answer is (3)

I.  Evaporate 50 mL of solution 
mmoles of glucose initially = 0.01 x 200 = 2 

Volume after evaporation= 200 - 50 = 150 mL 

${\mathrm{M}}_{\text{glucose }}=\frac{\text{ mmoles }}{{\mathrm{V}}_{\mathrm{mL}}}=\frac{2}{150}=0.013\mathrm{M}$
II. mmoles of glucose added = $\frac{0.180}{180}\times 1000=1$

Total mmoles of glucose = 2 + 1 = 3

Volume = 150 mL (after evaporation of 50 mL solution)

M_glucose = $\frac{3}{150}=0.02\mathrm{M}$

III. Add 50 mL of water:

New volume of solution = 200 + 50 = 250 mL

M_glucose =$\frac{2 \mathrm{mmol}}{250 \mathrm{mL} }=0.008\mathrm{M}$

Hence answer is (3)