Question #98df5

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Question #98df5

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Answer 1 · 2015-11-24T21:14:30

$11.26 g$ $11.26g$ of Glucose.

Explanation:

Glucose $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ has a molar mass of $180.16 \frac{g}{mol}$ $180.16 g/"mol"$ .

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

$C_{M} = \frac{n}{V} \Rightarrow n = C_{M} \times V = 0.25 \frac{mol}{L} \times 0.25 L = 0.0625 m o l$ $C_M=n/V=>n=C_MxxV=0.25"mol"/cancel(L)xx0.25cancel(L)=0.0625mol$

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

$m = n \times M M = 0.0625 m o l \times 180.16 \frac{g}{m o l} = 11.26 g$ $m=nxxMM=0.0625cancel(mol)xx180.16g/(cancel(mol))=11.26g$

Therefore, you will have to add $11.26 g$ $11.26g$ of glucose to a $250 m L$ $250mL$ -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.

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