A solution is made by dissolving 1.41 g glucose (C 6 H 12 O 6 ) in 100.0 g of wa

Question

A solution is made by dissolving 1.41 g glucose (C6H12O6) in 100.0 g of water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molality of glucose in this solution.

What is the amount of C6H12O6 present in moles?

Can anyone help me get started on this?

Do I use these equations?

Molarity= Moles of solute/liters of solution

Molaity(m)= moles of solute/ kilogram of solvent

Mass (wieght) %= mass of solute/mass of solution * 100%

Mole fraction= moles/total moles of solution

Thank you!

Explanation / Answer

Molarity = NO.of moles of solute×1000/V(ml)

No of moles of solute(Glucose)=1.41/180    ( Mol weight of Glucose=180)

                                                       =0.00783

Molarity of Glucose=0.00783×1000/101

                                  =0.07752

Mlality(m) = Moles of solute/kilogram of solvent

                    = 0.00783×1000/100

                    =0.0783

Mass(weight %)= Mass of solute(Glucose)×100/Mass of solution

                            =1.41×100/101

                            =1.396

Mole fraction=No of moles of solute(glucose)/Total no.of moles of solution

                          =    0.00783/0.00783+5.55

                          =0.00141

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