Question

STARTING AMOUNT

X
What quantity of moles of glucose (C,H,O) is in 0.500 L of a 0.400 M
solution?
ADO FACTOR
x( )
0.500 0.400
36.03
1.20 x 10 0.020
mol glucose g glucose
mL
ANSWER
6.022 x 10 180.18 7.53 x 10 0.200
1.25 2.08 x 1024
225
L
RESET
M glucose
2

Answer 1

Starting with the information provided:

You have a solution with a concentration of 0.400 M (moles per liter) of glucose (C6H12O6).

The volume of the solution is 0.500 L.

Let's break down the calculations step by step:

Moles of Glucose (C6H12O6):

Moles = Concentration (M) × Volume (L)

Moles = 0.400 mol/L × 0.500 L

Moles = 0.200 moles

Molar Mass of Glucose:

The molar mass of glucose (C6H12O6) is approximately 180.18 g/mol.

Mass of Glucose:

Mass = Moles × Molar Mass

Mass = 0.200 moles × 180.18 g/mol

Mass = 36.036 g (rounded to three decimal places)

Avogadro's Number:

Avogadro's number is approximately 6.022 × 10^23 particles/mol.

Number of Glucose Molecules:

Number of molecules = Moles × Avogadro's Number

Number of molecules = 0.200 moles × 6.022 × 10^23 molecules/mol

Number of molecules = 1.2044 × 10^23 molecules

Please note that the numbers in your calculations seem to be a mix of various units and factors. I've attempted to interpret and calculate based on the given context. If you have any specific questions about the calculations or need further clarification, please let me know!