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.
Answer:
When the pressure increases to 90.0 atm , the volume of the sample is 0.01467L
Explanation:
To answer the question, we note that
P₁ = 1.00 atm
V₁ = 1.32 L
P₂ = 90 atm.
According to Boyle's law, at constant temperature, the volume of gas is inversely proportional to its pressure
That is P₁V₁ = P₂V₂
Solving the above equation for V₂ we have
that is V₂ = 
= 
or 0.01467L
