Calculate molarity from mass and volume: A solution is prepared from `22.5 g` of glucose `(C_(6)H_(12)O_(6))` and enough water for a volume of `250 cm^(3)`. Find the molarity of the glucose. Strategy: We first convert mass glucose to moles, because the molarity equals moles of solute (giucose) divided by volume of solution in litres. To calculate moles, we must divide the given mass of glucose by its molar mass. Molar is gram atomic mass (if solute consists of atoms), gram molecular mass (if solute consists of molecules) or gram formula mass (if solute consista of formula units) Gram atomic mass is the mass in grams, numerically equal to the atomic mass expressed in amu. Gram molecular mass is the mass in grams, which is numerically equal to the molecular mass expressed in amu. Gram formula mass is the mass in grams, numerically equal to formula mass expressed in amu.

Calculate molarity from mass and volume: A solution is prepared from `

1.	Calculate molarity from mass and volume: A solution is prepared from `22.5 g` of glucose `(C_(6)H_(12)O_(6))` and enough water for a volume of `250 cm^(3)`. Find the molarity of the glucose. Strategy: We first convert mass glucose to moles, because the molarity equals moles of solute (giucose) divided by volume of solution in litres. To calculate moles, we must divide the given mass of glucose by its molar mass. Molar is gram atomic mass (if solute consists of atoms), gram molecular mass (if solute consists of molecules) or gram formula mass (if solute consista of formula units) Gram atomic mass is the mass in grams, numerically equal to the atomic mass expressed in amu. Gram molecular mass is the mass in grams, which is numerically equal to the molecular mass expressed in amu. Gram formula mass is the mass in grams, numerically equal to formula mass expressed in amu.
Answer» Glucose is a polar organic compound consisting of `C_(6)H_(12)O_(6)` molecules (180 u molecule -1). Thus, its molar mass will be expressed as gram molecular= `"Mass"/("Molar mass")` `(n_("glucose"))=(22.5 g "glucose")/(180 g "glucose"//"mole glucose")` =0.125 mol glucose Volume of the solution in liters `250 cm^(3)xx(1 L)/(1000 cm^(3))=0.25 L` solution Using equation `2.13` `M=n_("glucose")/V_(L)=(0.125 "mol glucose")/(0.25 L "solution")` `=0.5 mol L^(-1)` `=0.5 M`

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