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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.-mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0-mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL of the final solution

Answer :

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

[tex]\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M[/tex]

To calculate the molarity of the diluted solution, we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the concentrated glucose solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted glucose solution

We are given:

[tex]M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL[/tex]

Putting values in above equation, we get:

[tex]0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M[/tex]

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

[tex]0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g[/tex]

Hence, the mass of glucose in final solution is 0.420 grams

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