# Question #98df5

Nov 24, 2015

$11.26 g$ of Glucose.

#### Explanation:

Glucose ${C}_{6} {H}_{12} {O}_{6}$ has a molar mass of $180.16 \frac{g}{\text{mol}}$.

From the molarity (${C}_{M}$) and the volume ($V$) we can calculate the number of mole of glucose that will be present in the solution:

${C}_{M} = \frac{n}{V} \implies n = {C}_{M} \times V = 0.25 \frac{\text{mol}}{\cancel{L}} \times 0.25 \cancel{L} = 0.0625 m o l$

Using the number of mole expression $n = \frac{m}{M M}$ we can calculate the mass of glucose needed to be used:

$m = n \times M M = 0.0625 \cancel{m o l} \times 180.16 \frac{g}{\cancel{m o l}} = 11.26 g$

Therefore, you will have to add $11.26 g$ of glucose to a $250 m L$-volumetric flask then add water to the mark.

Here is a video that explains in details how you prepare a solution along with the calculations.

Lab Demonstration | Solution Preparation & Dilution.