Question

What mass of dextrose, C6H12O6 is dissolved in 325 mL of 0.258 M solution?

Answer 1

Answer on Question #76685 – Chemistry – Other

Task:

What mass of dextrose, C6H12O6 is dissolved in 325 mL of 0.258 M solution?

Solution:

We are given the concentration (0.258 M = 0.258 mol*L⁻¹).

Using a periodic table, we find the molar mass of C₆H₁₂O₆:

$$M \left(C _ {6} H _ {1 2} O _ {6}\right) = 6 * A r (C) + 1 2 * A r (H) + 6 * A r (O);$$ $$M \left(C _ {6} H _ {1 2} O _ {6}\right) = 6 * 1 2. 0 1 0 7 + 1 2 * 1. 0 0 7 9 4 + 6 * 1 5. 9 9 9 = 1 8 0. 1 5 3 4 8 \frac {g}{m o l}.$$ $$M \left(C _ {6} H _ {1 2} O _ {6}\right) \approx 1 8 0. 1 5 \frac {g}{m o l}$$

The molar mass of C₆H₁₂O₆ is 180.15 g·mol⁻¹.

We will convert 325 mL into 0.325 L.

$$C (X) = \frac {n (X)}{V (X)} = \frac {m (X)}{M (X) * V (X)} \Rightarrow m (X) = C (X) * M (X) * V (X)$$

Then,

$$m \left(\mathrm {C} _ {6} \mathrm {H} _ {1 2} \mathrm {O} _ {6}\right) = C \left(\mathrm {C} _ {6} \mathrm {H} _ {1 2} \mathrm {O} _ {6}\right) * M \left(\mathrm {C} _ {6} \mathrm {H} _ {1 2} \mathrm {O} _ {6}\right) * V \left(\mathrm {C} _ {6} \mathrm {H} _ {1 2} \mathrm {O} _ {6}\right);$$ $$m \left(\mathrm {C} _ {6} \mathrm {H} _ {1 2} \mathrm {O} _ {6}\right) = 0. 2 5 8 \frac {\text {mol}}{L} * 1 8 0. 1 5 \frac {\mathrm {g}}{\text {mol}} * 0. 3 2 5 L = 1 5. 1 0 5 6 \mathrm {g};$$ $$m \left(\mathrm {C} _ {6} \mathrm {H} _ {1 2} \mathrm {O} _ {6}\right) \approx 1 5. 1 \mathrm {g}$$

Answer: 15.1 g of dextrose will be required.

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