Molarity Problem; pre-solved problem, wondering where certain numbers come from?

M

Mama Tooty

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Molarity
A 100.5 mL intravenous (IV) solution contains 5.10g of glucose (C6H12O6). What is the Molarity of this solution? The molar mass of glucose is 180.16 g/mol
Known
Mass of solute = 5.10 g C6H12O6
Molar Mass =180.16 g/mol
Volume of Solution = 100.5 mL
Unknown
Solution concentration = ? M

Calculate the number of moles of C6H12O6
(5.10 g C6H12O6)(1 mol C6H12O6)(180.16 g C6H12O6)
= 0.0283 mol C6H12O6
Convert the volume of H2O to liters
(100.5 mL solution)(1 L / 1000mL) = 0.1005 L solution
Solve for molartiy
M = Moles of solute/ liters of solutions
M = (0.0283 mol C6H12O6 / 0.1005 L solution)
M = (0.0282 mol C6H12O6 / 1 L Solution) = 0.282 M
This is a problem already solved for me out of the book, I was wondering where they came up with the last part of the equation: M = (0.0282 mol C6H12O6 / 1 L Solution) = 0.282 M
Could someone tell me how to find these numbers.
 
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