there are for given choices
a. 1.44 x 10^22
b. 1.44 x 10^25
c. 2.41 x 10^21
d. 2.41 x 10^24
the correct answer is A but i dont understand how to get it. i tried converting mmol to mol and got 0.004 mol C6H6O6 and then coverted that to 0.024 mol C by multiplying it by 6 mol C. I then converted mol C to grams carbon, getting .288 g C. Then i used 1.66053873 x 10^ -24 to convert it to amu, getting 1.73 x 10^23 which isn't one of the answers. thank you
2006-12-12 08:13:23 · 7 answers · asked by E.T.01 5 in Science & Mathematics ➔ Chemistry
number of glucose atoms in 1 mol = avagadros constant = 6.023 x 10^23
number of glucose atoms in 1mmol = avagado / 1000 = 6.023x10^20
number of glucose atoms in 4mmol = 4 x (6.023x10^20) = 2.41 x 10^21.
6 carbon atoms per glucose atom so answer = 6 x 2.41 x 10^21
= 1.44 x 10^22. Answer A - I am a geek!
2006-12-12 11:52:56 · answer #1 · answered by chem geek 1 · 1⤊ 0⤋
Using the Avogadro Constant
6.022045 x 10^23
This is the number of molecules in a mole.
In your glucose molecule you have 6 carbon atoms. So you must multiply the Avogadro Constant by 6 to calculate the number of carbon atoms in one mole of C6H6O6
So 6 x 6.022045 x 10^23 = 3.613227 x 10^24 carbon atoms in one mole of glucose.
However, you only have 4.00 mmol of glucose. That is 4 thousandths of one mole. or 4.0 x 10 ^-3.
So multiply 3.613227 x 10^24 x 4.0 x 10^-3 = 1.4452908 x 10^22
or to 2 dp 1.44 x 10^22 as required.
2006-12-12 08:55:13 · answer #2 · answered by lenpol7 7 · 2⤊ 0⤋
1 mol of any molecules contains 6.02*10^23 molecules
So 4mmol = 4*10-3 mol is equivalent to 6.02*10^23 * 4*10^-3 = 2.408*10^21 molecules
Each molecule contains 6 C atoms.
So 2.408*10^21 molecules contain 2.408*10^21 * 6 = 1.44*10^22 C atoms
2006-12-12 08:19:59 · answer #3 · answered by claudeaf 3 · 1⤊ 0⤋
Avogadro's number = 6.0221415 x 10^23
So there are 6.0221415 x 10^23 atoms in 1 mole of substance.
Number of C atoms in glucose
= (6 x 4 x 10^-3) x (6.0221415 x 10^23)
= 14,453,139,600,000,000,000,000
= approximately 1.44 x 10^22
2006-12-12 10:46:45 · answer #4 · answered by Kemmy 6 · 1⤊ 0⤋
Let G mean Glucose. Let C mean carbon. Given: 4.00mmolG To find: atomsC
4.00mmolG x (1x10^-3molG)/(1mmolG) x (6.02x10^23moleculesG)/(1molG) x (6atomsC)/(1moleculeG) = atomsC
In the first step, mmolG cancel to leave molG. In the second step, molG cancel to leave moleculesG. In the third step, moleculesG cancel to leave atomsC.
2006-12-12 08:25:46 · answer #5 · answered by steve_geo1 7 · 1⤊ 0⤋
6.022x10^23 x 0.004 x 6
2006-12-12 08:16:51 · answer #6 · answered by Anonymous · 1⤊ 0⤋
you have to use avagadros content 6.022x10^23
2006-12-12 09:01:30 · answer #7 · answered by Michael D 6 · 1⤊ 0⤋