Question #98df5

Question

4133 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-24T21:14:30

$11.26g$ of Glucose.

Glucose $C_6H_12O_6$ has a molar mass of $180.16 g/"mol"$ .

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:

$C_M=n/V=>n=C_MxxV=0.25"mol"/cancel(L)xx0.25cancel(L)=0.0625mol$

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

$m=nxxMM=0.0625cancel(mol)xx180.16g/(cancel(mol))=11.26g$

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

Please Like, Share and Subscribe.