Question

Question #98df5

Answer 1

`11.26g` of Glucose.

Explanation:

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.

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