Question

How many molecules of glucose,

C6H12O6, are present in 240.0 grams?
[?] x 10 molecules C6H12O6
Enter the coefficient in the green blank and the
exponent in the yellow blank. Report your answer
to the appropriate number of significant figures!
Coefficient (green)
Exponent (Yellow)
Enter

Answer 1

The molecular weight of glucose (C6H12O6) is 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol.

So, the number of moles of glucose present in 240.0 g can be calculated as follows:

n = m/M = 240.0 g / 180.18 g/mol = 1.33 mol

And the number of molecules of glucose can be calculated as follows:

N = n x Avogadro's constant = 1.33 mol x 6.022 x 10^23 molecules/mol = 7.98 x 10^23 molecules

Therefore, the number of molecules of glucose present in 240.0 g is 7.98 x 10^23 molecules, rounded to the nearest whole number.

So, the coefficient is 7.98 and the exponent is 23.

Step-by-step explanation: