
Which of the following aqueous solutions has maximum freezing point?
(A) $ 0.01\,M\,NaCl $
(B) $ 0.005M{{ }}{C_2}{H_5}OH $
(C) $ 0.005M{{ }}MgC{l_2} $
(D) $ 0.01M{{ }}MgS{O_4} $
Answer
545.1k+ views
Hint: Depression in freezing point is a colligative property of a solution i.e., it depends on the moles of solute irrespective of the nature of particles present in the solution. We need to calculate the value of the van't Hoff factor of the given compounds to check which has the maximum freezing point.
Complete Step-by-step Answer
When ionic compounds are dissolved in water they dissociate into their respective anions and cations, so the van't Hoff factor $ 'i' $ has to be taken into consideration. This will increase the net particle concentration in comparison to the covalent compounds.
The Depression in freezing point is given by the expression;
$ \Delta {T_f} = i \times {K_f} \times m\; $
Where, $ i = $ Van’t Hoff factor OR No. of dissolved particles
$ m = $ The molality of the solution
$ {K_f} = $ Molal depression constant
Now we will calculate the value of van’t Hoff factor for the given compounds.
For $ NaCl $ ; It will dissociate into $ N{a^ + } $ cation and $ C{l^ - } $ anion i.e. $ NaCl\; \to \;N{a^ + } + \;{{ }}C{l^ - } $
Thus the value of $ i = 2 $ and its effective concentration will be $ 0.02\,M $
For $ {C_2}{H_5}OH $ ; it will dissociate into $ {C_2}{H_5}OH \to {C_2}H_5^ - \, + O{H^ - } $ . Thus the value of $ i = 2 $ and its effective concentration will be $ 0.01\,\,M $
For $ MgC{l_2} $ ; It will dissociate into $ MgC{l_2} \to M{g^ + } + 2C{l^ - } $ . Thus the value of $ i = 3 $ and its effective concentration will be $ 0.01\,5\,M $
For $ MgS{O_4} $ ; It will dissociate into $ MgS{O_4} \to M{g^ + } + S{{O_4}^2}^- $ . Thus the value of $ i = 2 $ and its effective concentration will be $ 0.02\,M $ .
Now, from the above compounds $ {C_2}{H_5}OH $ has the least concentration. It means its depression in the freezing point will be minimum which means It will have the highest freezing point.
Hence, Option (B) is the correct option.
Note
The Van’t Hoff factor is defined as the ratio of moles of particles formed when a substance is dissolved to the moles of the substance by mass. The more the number of moles of solute the more will be the Van’t Hoff factor and more will be the value of the depression in the freezing point of the solvent.
Complete Step-by-step Answer
When ionic compounds are dissolved in water they dissociate into their respective anions and cations, so the van't Hoff factor $ 'i' $ has to be taken into consideration. This will increase the net particle concentration in comparison to the covalent compounds.
The Depression in freezing point is given by the expression;
$ \Delta {T_f} = i \times {K_f} \times m\; $
Where, $ i = $ Van’t Hoff factor OR No. of dissolved particles
$ m = $ The molality of the solution
$ {K_f} = $ Molal depression constant
Now we will calculate the value of van’t Hoff factor for the given compounds.
For $ NaCl $ ; It will dissociate into $ N{a^ + } $ cation and $ C{l^ - } $ anion i.e. $ NaCl\; \to \;N{a^ + } + \;{{ }}C{l^ - } $
Thus the value of $ i = 2 $ and its effective concentration will be $ 0.02\,M $
For $ {C_2}{H_5}OH $ ; it will dissociate into $ {C_2}{H_5}OH \to {C_2}H_5^ - \, + O{H^ - } $ . Thus the value of $ i = 2 $ and its effective concentration will be $ 0.01\,\,M $
For $ MgC{l_2} $ ; It will dissociate into $ MgC{l_2} \to M{g^ + } + 2C{l^ - } $ . Thus the value of $ i = 3 $ and its effective concentration will be $ 0.01\,5\,M $
For $ MgS{O_4} $ ; It will dissociate into $ MgS{O_4} \to M{g^ + } + S{{O_4}^2}^- $ . Thus the value of $ i = 2 $ and its effective concentration will be $ 0.02\,M $ .
Now, from the above compounds $ {C_2}{H_5}OH $ has the least concentration. It means its depression in the freezing point will be minimum which means It will have the highest freezing point.
Hence, Option (B) is the correct option.
Note
The Van’t Hoff factor is defined as the ratio of moles of particles formed when a substance is dissolved to the moles of the substance by mass. The more the number of moles of solute the more will be the Van’t Hoff factor and more will be the value of the depression in the freezing point of the solvent.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

