This is false. An alcohol does indeed have a polar C-O single bond, but what we should really be focusing on is the extraordinarily polar O-H single bond. When oxygen, fluorine, or nitrogen is bound to a hydrogen atom, there is a small (but not negligible) charge separation, where the eletronegative N, O, or F has a partial negative charge, and the H has a partial positive charge. Water has two O-H single bonds in it (structure is H-O-H). The partially negative charge on the O of the water molecule (specifically around the lone pair) can become attracted either a neighboring water molecule's partially positive H atom, or an alcohol's partially positive H atom. This is weak (and partially covalent) attraction is called a hydrogen bond. This is stronger than a typical dipole-dipole attraction (as would be seen between neighboring C-O single bonds), and much stronger than dispersion forces (between any two atoms). When the solvent (water) and the solute (the alcohol) both exhibit similar intermolecular forces (hydrogen bonding being the most important in this case), they can mix completely in all proportions (i.e. they are miscible) in water.
Moles Cu+2 = M * V
= 0.05 L * 0.011 m
= 0.00055 moles
when the molar ratio of Cu2+: EDTA = 1:1 so moles od EDTa also =0.00055 moles
and when the Molarity of EDTa = 0.0630 M
∴ Volume of EDTA = moles / Molarity
= 0.00055 / 0.0630
= 0.0087 L = 8.7 L
