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Delvig [45]
2 years ago
12

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 i

n 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.
Chemistry
1 answer:
BARSIC [14]2 years ago
3 0

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.

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