what is molarity of glucose solution G? a stock solution is prepared by dissolving .1802 g of glucose in water using a 1 L volumetric flask (precision is plus minus .1 ml). a dilution (solution A) is made by taking 1.00 ml of stock solution and dilution to 1 L is repeated six more times using the same glassware to prepare solution B through G. what is the molarity of glucose solution G?

A stock solution of glucose is prepared by dissolving of 0.1802g of glucose in water using a 1-L volumetric flask (precision is (+/-) 0.1 mL ). A dilution (solution "A") is made by taking 1.00mL of the stock solution (using a pipet) and diluting to 1L with water (using a volumetric flask). This dilution procedure (dilute a 1-mL aliquot to 1L) is repeated six more times using the same glassware to prepare solution "B" through "G". what is the molarity of glucose solution G?

(end of the question)

Solution to the problem:

1. Calculate the molarity of the stock solution:

[tex]\text{Molarity}=\frac{\text{moles of solute}}{\text{volume of solution in liters}}[/tex]

[tex]\text{moles of solute}=\frac{mass in grams}{molar mass}[/tex]

Molar mass of glucose: 180.156 g/mol (from internet)

[tex]\text{number of moles}=\frac{0.1802}{180.156g/mol}=0.001000mol[/tex]

[tex]Molarity=0.0010000mol/1liter=0.0010000M[/tex]

[tex]Molarity=1.0000\times 10^{-3}M[/tex]

2. Calculate the molarity of the first dilution (solution A)

[tex]Molarity_G\times Volume_G=Molarity_A\times Volume_A[/tex]

[tex]Molarity_A=1.0000\times 10^{-3}M\times (1.00ml\times 1liter/1,000ml)/1iter\\\\ Molarity_A=1.0000\times 10^{-6}M[/tex]

The solution was diluted by a factor of [tex]1.0000\times 10^{-3}/1.0000\times 10^{-6}=10^3[/tex]

You can realize that every time that you dilute the solution, its molarity will dcrease by a factor of 10³.

Then, the dilutions is made 6 more times, you can divide the molarity of solution A by [tex]{(10^3)}^6=10^{18}[/tex] .

Hence, the molarity of solution G is:

[tex]Molarity_G=10^{-6}/10^{18}=10^{-24}M[/tex]