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Since toluene is a non-polar compound i.e., it doesn't have any polar group so it is insoluble in water. (because water is a polar compound)
We know the fact that like dissolves in like. Since phenol is had both polar and non-polar group so it is partially soluble in water.
The order will  be :           Cyclohexane > CH3CN > CH3OH > KCl In this, we have used the fact that like dissolves like. Since cyclohexane is an alkane so its solubility will be maximum. 
According to given conditions we have same solution under same temperature. So we can write :                                                                                                              So, if we put all the given values in above equation, we get                                                  or                                               Hence the required concentration is...
In this question we will use the formula :                                                                Firstly for compound AB2  :-                                                    or                                                        Similarly for compound AB4 :-                                                    If we assume atomic weight of element A to be x and of element B to be...
It is given that freezing point of pure water is 273.15 K. So, elevation of freezing point = 273.15 - 271 = 2.15 K  solution means 5 g solute in 95 g of water. Moles of cane sugar :                                         Molality :                                           We also know that -               or                                           Now we will use the above procedure for...
In the previous part we have calculated the value of molar mass the Raoul's law equation. We had :-                               Putting M = 23 u in the above equation we get,                               or                          Thus vapour pressure of water = 3.53 kPa.
In this question we will find molar mass of solute by using Raoult's law . Let the molar mass of solute is M. Initially we have 30 g solute and 90 g water. Moles of water :                               By Raoult's law we have :-                                                 or                                         or                                                                          ...
Let the initial vapour pressure of octane = . After adding solute to octane, the vapour pressure becomes :                                                                      Moles of octane :                                                    Using Raoult's law we get :                                             or                                      or                                   ...
It is asked the vapour pressure of 1 molal solution which means 1 mol of solute in 1000 g H2O. Moles in 1000g of water = 55.55 mol.         (Since the molecular weight of H2O is 18) Mole fraction of solute :                                                   Applying the equation :                                            or                                        or                           ...
Vapour pressure of heptane =     and vapour pressure of octane =    Firstly we will find moles of heptane and octane so that we can find vapour pressure of each. Molar mass of heptane =  7(12) + 16(1)  =  100 unit. and molar mass of octane =  8(12) + 18(1) = 114 unit. So moles of heptane :                                           and moles of octane :                                      ...
It is given that  of aq. solution. This means 2 g of non-volatile solute in 98 g of H2O. Also the vapour of water at normal boiling point = 1.013 bar. Using Raoult's law :                                        So we get :                                        Thus the molar mass of non-volatile solute is 41.35 unit.
Positive and negative deviation:- A non-ideal solution is defined as a solution which does not obey Raoult’s law over the entire range of concentration i.e.,      and . The vapour pressure of these solutions is either higher or lower than that expected by Raoult’s law. If vapour pressure is higher, the solution shows a positive deviation and if it is lower, it shows a negative deviation...
Using Henry's Law we can write,                                                               Putting value in this equation, we get :                                                So, the magnitude of k is   . Now, we will again use the above equation for  . So the required partial pressure is :-                                              or                                                 ...
According to Henry's law at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of solution or liquid. i.e.,   p = khx, Here kh is Henry’s law constant. Some of its applications are as follows:- (a)  We can increase the solubility of CO2 in soft drinks, the bottle is sealed under high pressure. (b)  To...
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