How many glucose molecules are included in that 200 mL sample?

Question

You have a freshly prepared 1 M solution of glucose in water. You carefully pour out a 200 mL sample of that solution. 

Answer 1

 

Maybe you can apply the dilution formula interpreting the data this way

 

The Molarity of the initial solution is equal to 1, M=1, for V should be equal to 1 L , being M=n/V as mol/L

 

The final solution is the one with the volume of the 200 mL  of solution, and this is the one you must calculate M from the dilution formula and from where the number of moles,n, of the final solution to obtain the number of molecules,N, of glucose of the final solution.

 

It is a longer process, but another way to explain how to arrive to the result.

 

Mi=1  Vi= 1L=         Mf=x  Vf= 200mL= 0.2 L

 

The formula is given by this product :

 

             MiVi= MfVf

 

In the case,  1*1= x*0.2

 

From where Mf, the unknown Molarity will be:

 

X= Mf= MiVi/Vf= 1/0.2= 5 mol/L, a solution more concentrate, so with more solute compared to the previous

 

Mf= 5 mol/L

 

 From M=n/V 

 

n = MV

 

nf= 5* 0.2 = 1 mol

 

Finally the number of molecules of final solution, N

 

N= n*L   (L= Avogadro's number= 6.022*10²³⁾

 

N = 1* 6.022*10²³

 

N= 6.022*10²³

 

The final solution, made of 200 mL of sample, equal to 0.2 L, will have a number of Avogadro's of molecules of glucose per liter of solution.

 

 

 

 

 

 

Answer 2

I think we need more information, such as the original volume of the 1 M (I assume that means 1 mole) solution. Suppose the volume before pouring is V.  If there is 1 mole dissolved in V, then there are 6.02 x 10²³ glucose molecules in V.  Then the number of glucose molecules in the 200 mL sample is

 

N = (200 mL / V) * (6.02 x 10²³)