Question

A 2.550 x 10^−2 M solution of glycerol (C3H8O3) in water is at 20.0°C. The sample was created by dissolving a sample of C3H8O3 in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 998.9 mL . The density of water at 20.0°C is 0.9982 g/mL.
a. Calculate the molality of the glycerol solution.

Answer 1

Answer:

$2.557\times 10^{-2}\ molal.$

Explanation:

Given , molarity of glycerol= $2.55\times 10^-^2\ M.$

Volume= 1 L.

Therefore, No of moles of glycerol= $molarity\times volume\ in\ liters=2.55\times 10^-^2\ moles.$

Now, volume of water needed, V=998.8 mL.

Density is given as= 0.9982 g/mL.

Therefore, mass of water = $volume\times density=998.8\times 0.9982=997\ g=0.997\ kg$

Now, molality=$\dfrac{no\ of\ moles}{mass\ of\ solvent\ in\ kg}=\dfrac{2.55\times 10^-^2\ }{0.997}=0.02557\ molal=2.557\times 10^{-2}\ molal.$

Hence, this is the required solution.