Question
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..
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..
Answers
bobpursley
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.
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.
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