Question

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 1

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

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}` .........(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:

`\text{Initial molarity of glucose}=(10.5* 1000)/(180.16* 100)\\\\\text{Initial molarity of glucose}=0.583M`

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

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the molarity and volume of the concentrated glucose solution

`M_2\text{ and }V_2` are the molarity and volume of diluted glucose solution

We are given:

`M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL`

Putting values in above equation, we get:

`0.583* 20=M_2* 500\\\\M_2=(0.583* 20)/(500)=0.0233M`

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:

`0.0233=\frac{\text{Mass of glucose in final solution}* 1000}{180.16* 100}\\\\\text{Mass of glucose in final solution}=(0.0233* 180.16* 100)/(1000)=0.420g`

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