# The molar mass of glucose is 180.16g/mol

A. How much glucose is needed to make 500 ml of a .44M (.44mol/1L) solution?
.44M/L=180.16x.44/100ml=79.27/1000=.07927x500=39.64g/mol/550ml?
B What volume of the .44M solution is needed to prepare 1L of a 5% glucose solution? Don't know where to start with this one..

## molarity=mass/molmass * 1/volumeinliters

mass=molarity*molmass*volumeinliters
= .44*180*.5 grams you do it

b. 5 precent by mass is what molarity?
one liter of 5 percent is /1000g ignoring that the dilute solution is slightly greater than 1000g per liter.
massin5 percent=50 grams.
molarity of that then= 50/180*1/1 = .28M

so you want to dilute it .44/.28 times?
.44/.28=1.57

which means one part original solution, .57 psrts water added.
What is one part? 1L/1.57 = 636ml of the stock solution

check my math.

## To calculate the volume of the .44M solution needed to prepare 1L of a 5% glucose solution, follow these steps:

Step 1: Calculate the amount of glucose needed in 1L of the 5% solution.
5% of 1L = 0.05 x 1000ml = 50g

Step 2: Calculate the moles of glucose needed.
Molar mass of glucose = 180.16g/mol
Moles of glucose = mass / molar mass
Moles of glucose = 50g / 180.16g/mol ≈ 0.277moles

Step 3: Calculate the volume of the .44M solution that contains 0.277moles of glucose.
Volume (L) = moles / molarity
Volume (L) = 0.277moles / 0.44mol/L ≈ 0.630L

Therefore, approximately 0.630L (or 630ml) of the .44M solution is needed to prepare 1L of a 5% glucose solution.

## To find the volume of the .44M solution needed to prepare 1L of a 5% glucose solution, you can follow these steps:

Step 1: Determine the molar mass of glucose.
Given that the molar mass of glucose is 180.16g/mol.

Step 2: Calculate the mass of glucose needed to make the 5% solution.
The term "5% glucose solution" means that there are 5g of glucose in 100ml of solution. Since we are preparing 1L (1000ml) of the solution, we can calculate the mass of glucose needed as:
Mass of glucose needed = (5g/100ml) x 1000 ml = 50g

Step 3: Calculate the moles of glucose needed.
Moles of glucose = Mass of glucose needed / Molar mass of glucose
Moles of glucose = 50g / 180.16g/mol = 0.277mol

Step 4: Use the molarity formula to find the volume of the .44M solution needed.
Molarity (M) = Moles (mol) / Volume (L)
0.44M = 0.277mol / Volume
Volume = 0.277mol / 0.44M
Volume = 0.630 L or 630 ml

So, to prepare 1L of a 5% glucose solution, you would need approximately 630 ml of the .44M solution.