Question

3.25 ×10^26 molecules of C6H12O6 to grams

Answer 1

Step-by-step explanation:

`N=n* N_A`

Where:

N = Number of particles / atoms/ molecules

n = Number of moles

`N_A=6.022* 10^(23) mol^(-1)` = Avogadro's number

We have:

Molecules of glucose,
`C_6H_(12)O_6=3.25* 10^(26)`

Moles of glucose = n =?

`n=(N)/(N_A)=(3.25* 10^(26))/(6.022* 10^(23) mol^(-1))`

n = 539.69 mol

Mass of 539.69 moles of glucose :

= 539.69 mol × 180 g/mol =97,144.2 g

Answer 2

Answer: -

9270 grams

Explanation: -

Number of molecules of C₆H₁₂O₆ = 3.25 x 10 ²⁶

According to Avogadro's law,

6.02 x 10 ²³ molecules of C₆H₁₂O₆ are present in 1 mole of C₆H₁₂O₆

3.25 x 10 ²⁶ molecules of C₆H₁₂O₆ are present in
`(1 mole C6H12O6 x 3.25 x 10 26 molecules)/(6.02 x 10 23 molecules)` of C₆H₁₂O₆

= 54 mol of C₆H₁₂O₆

Molar mass of C₆H₁₂O₆ = 12 x 6 + 1 x 12 + 16 x 6 = 180g /mol

Mass of C₆H₁₂O₆ = 54 mol x 180g /mol

= 9270 g