Question

(LO 4E, 4F) A student placed 11.0 g of glucose (C6H12O6) in a volumetric flask, 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 0.500L. What is the concentration of the dilute solution?

Answer 1

Answer:

0.244 M

Explanation:

Calculation of the moles of glucose as:-

Mass = 11.0 g

Molar mass of glucose  = 180.156 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{11.0\ g}{180.156\ g/mol}$

$Moles_{glucose}= 0.0611\ mol$

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Given that:- Volume added = 100 mL = 0.1 L

So, Molarity of the glucose solution, $Molarity=\frac{0.0611}{0.1}\ M=0.611\ M$

Considering

$Molarity_{diluted\ solution}\times Volume_{diluted\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given  that:

$Molarity_{diluted\ solution}=?$

$Volume_{diluted\ solution}=0.500\ L$

$Volume_{stock\ solution}=20.0\ mL=0.2\ L$

$Molarity_{stock\ solution}=0.611\ M$

So,  

$Molarity_{diluted\ solution}\times 0.5=0.611\ M\times 0.2$

$Molarity_{diluted\ solution}=\frac{0.611\ M\times 0.2}{0.5}=0.244\ M$

The molarity of the diluted solution is:- 0.244 M