Question

Which of the following is soluble in water?
(A) $C{{S}_{2}}$
(B) ${{C}_{2}}{{H}_{5}}OH$
(C) $CC{{l}_{4}}$
(D) $CHC{{l}_{3}}$

Answer 1

Hint: To answer this question, we should first check the polarity of the given options. We should know one concept that dissolves like.

Step by step answer:
We should know to predict solubility of compounds. To determine the solubility characteristics of some simple organic molecules and to look at relationships between the solubility properties of an organic molecule we should know about some properties.
We should know that solubility is controlled by the energy balance of intermolecular forces between solute-solute, solvent-solvent and solute-solvent molecules. In a very simple way, we can say that like dissolves like. This is the simplest rule to first check solubility of compounds. This simplest rule is based on polarity of solute and solvent. We should note that polar molecules dissolve in polar solvents and also we can say that non-polar molecules dissolve in nonpolar solvents.
So, to check the solubility of each option, we will first check the polarity of each option.
We should first know about water polarity. We should know that two hydrogen atoms and one oxygen atom within water molecules (${{H}_{2}}O$) form polar covalent bonds. This is because; the polarity of water creates a slightly positive charge on hydrogen and a slightly negative charge on oxygen, contributing to water's properties of attraction. So, we can say that water is polar. Now we have to find polar compounds in given options.
In the first option that is$C{{S}_{2}}$, we know that this is non-polar. So, it will not dissolve in water.
In the second option of ethyl alcohol we can say that as water is a polar solvent and has hydrogen bonding, so for ethyl alcohol compounds to dissolve in water, it must possess polarity or H-Bond. Ethyl alcohol shows presence of hydrogen bonding as well as it is polar due to high electronegativity of oxygen; due to this it is soluble in water.
In the third option, $CC{{l}_{4}}$ is non-polar. The tetrahedral symmetry of the molecule means any bond dipoles (C-Cl) will all cancel out. So, from this we can say that it will not be soluble in water.
In the fourth option, we can say that chloroform is slightly soluble in water. We should know that chloroform is a slightly polar compound. But it cannot make any strong bond (like hydrogen bond) with water. Due to lack of strong interaction with water its solubility in water is very less.
So, from the above discussion, we can say that option B is correct. Ethyl alcohol present in option B is soluble in water.

Note: We should know more about solubility. We should know that the solubility of a substance depends on the physical and chemical properties of the solute and solvent as well as on temperature, pressure and presence of other chemicals (including changes to the pH) of the solution.