A $4.00 m$ aqueous solution of glucose (molar mass $180 g mo l^{- 1}$ ) has density $1.20 g m L^{- 1}$ . Calculate its molarity in $mol L^{- 1}$ to two decimal places.

Numerical answer type, answer in mol/L, tolerance 0.01.

Answer: 2.79 mol/L

Take

1 kg

of water with

4.00 mol

glucose: solute mass

= 720 g

, solution mass

= 1720 g

. Volume

= 1720/1.20 = 1433.3 mL = 1.4333 L

. Molarity

= 4.00/1.4333 = 2.791 mol L^{- 1}