Question

What is the molarity of an aqueous solution containing 36.0 grams of fructose (C6H12O6) in 50.0 mL of solution? (C = 12.01 amu, H = 1.01 amu, O = 16.00 amu)

To make sure you are doing this one correct, determine all of the following:

mols = Blank 1 mols

L = Blank 2 L

M = Blank 3 M

Answer 1

1)

Molar mass of C6H12O6 = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass of C6H12O6 = 36.0 g

we have below equation to be used:

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(36.0 g)/(180.156 g/mol)

= 0.1998 mol

Answer: 0.1998 mol

2)

volume , V = 50.0 mL

= 5.00*10^-2 L

Answer: 5.00*10^-2 L

3)

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.1998/5*10^-2

= 3.997 M

Answer: 3.997 M