Question

A solution contains 90g of H20 6.4g of methanol and 18.4 g of glycerol

Answer 1

Hello there
The solution contains 90 g of H2O,so moles of H2O=given mass/molar mass
                                                                                  =90/18=5
                            moles of Methanol(CH3OH)=given mass/molar mass
                                                                        =6.4/32=0.2 mol
                            moles of glycerol(C3H8O3)=given mass/molar mass
                                                                       =18.4/92=0.2 mol
Now mole fraction of water=moles of H2O/moles of (H2O+CH3OH+C3H8O3)
                                          =5(5+0.2+0.2)=5/5.4=0.925
mole fraction of methanol
=moles of CH3OH/moles of (H2O+CH3OH+C3H8O3)=0.2/(5+0.2+0.2)=
                                                                                        0.037
mole fraction of glycerol=
moles of C3H8O3/moles of (H2O+CH3OH+C3H8O3)=0.2/(5+0.2+0.2)=
                                                                                         0.037 
Hope it helps.

Answer 2

Answer:

0.037

Explanation: Hello

The problem given us

Mass of solvent(H20)=90g

Mass of solute1(methanol)=6.4g

Mass of solute2(glycerol)=18.4g

We know that no.of mole of a chemical =mass of chemical/molar mass of chemical

So no.of moles of glycerol=18.4/92=0.2

No of moles of water=90/18=5

No.ofmoles of methanol=6.4/32=0.2

So mole fraction of glycerol=0.2/0.2+5+0.2=0.037(ans)