
The vapour pressure of pure water is $ 23.8{\text{ }}mm $ $ Hg $ at $ {25^o}C $ . What is the vapour pressure of $ 2.50{\text{ }}molal $ $ {C_6}{H_{12}}{O_6} $ .
Answer
475.5k+ views
Hint :Vapour pressure is actually a measure of the tendency of a substance to change into a gaseous or a vapour state, and to be noted that it increases with temperature.
Complete Step By Step Answer:
According to Raoult's law, the vapour pressure of a component at a given temperature equals the mole fraction of that component times the vapour pressure of that component in a pure state.
$ {P_{solution}} = {X_{solvent}}.P_{solvent}^o $
$ {P_{solution}} $ =Vapour pressure of solution (Here, solution is $ {C_6}{H_{12}}{O_6} $ )
$ {X_{solvent}} $ =mole fraction of solvent (Here, solvent is water)
$ {P^o}_{solvent} $ =Vapour pressure of pure solvent (Here solvent is water)
(And, mole fraction refers to the ratio of the number of moles linked to one component in a solution or mixture to the total number of moles).
In the question we are given with the following information:
$ P_{water}^o = 23.88mmHg $
$ Molarity{\text{ }}of{\text{ }}solution = 2.5{\text{ }}molal $
Number of mole of $ {C_6}{H_{12}}{O_6} = 2.50 $
Let us assume amount of water $ = 1kg $
$ M.W{\text{ }}of{\text{ }}water = 2 + 16 = 18 $
Thus number of moles i.e. n can be calculated as:
$ n = \dfrac{m}{M} $ where, m is the given mass
Using this formula, we will calculate the number of moles of water as $ \dfrac{{1000}}{{18}} $
So, total number of moles = $ \dfrac{{1000}}{{18}} + 2.5 $
Now, $ {X_{water}} = \dfrac{{\dfrac{{1000}}{{18}}}}{{\dfrac{{1000}}{{18}} + 2.5}} = \dfrac{{1000}}{{1045}} $
Now, substituting the values, we will calculate the vapour pressure of $ 2.50{\text{ }}molal $ $ {C_6}{H_{12}}{O_6} $ as stated below:
$ {P_{{C_6}{H_{12}}{O_6}}} = \dfrac{{1000}}{{1045}} \times 23.8 = 22.775{\text{ }}mmHg $
Hence, the vapour pressure of $ 2.50{\text{ }}molal $ $ {C_6}{H_{12}}{O_6} $ is $ 22.775{\text{ }}mmHg $ .
Note :
Molality, a measure of concentration, is especially used when we have to study the characteristics of solutions that are related to vapor pressure as well as alterations in temperature. In such cases, molality is preferred because its value does not vary with alteration in temperature. As in other measures of concentration, volume of solution is used which depends slightly upon temperature.
Complete Step By Step Answer:
According to Raoult's law, the vapour pressure of a component at a given temperature equals the mole fraction of that component times the vapour pressure of that component in a pure state.
$ {P_{solution}} = {X_{solvent}}.P_{solvent}^o $
$ {P_{solution}} $ =Vapour pressure of solution (Here, solution is $ {C_6}{H_{12}}{O_6} $ )
$ {X_{solvent}} $ =mole fraction of solvent (Here, solvent is water)
$ {P^o}_{solvent} $ =Vapour pressure of pure solvent (Here solvent is water)
(And, mole fraction refers to the ratio of the number of moles linked to one component in a solution or mixture to the total number of moles).
In the question we are given with the following information:
$ P_{water}^o = 23.88mmHg $
$ Molarity{\text{ }}of{\text{ }}solution = 2.5{\text{ }}molal $
Number of mole of $ {C_6}{H_{12}}{O_6} = 2.50 $
Let us assume amount of water $ = 1kg $
$ M.W{\text{ }}of{\text{ }}water = 2 + 16 = 18 $
Thus number of moles i.e. n can be calculated as:
$ n = \dfrac{m}{M} $ where, m is the given mass
Using this formula, we will calculate the number of moles of water as $ \dfrac{{1000}}{{18}} $
So, total number of moles = $ \dfrac{{1000}}{{18}} + 2.5 $
Now, $ {X_{water}} = \dfrac{{\dfrac{{1000}}{{18}}}}{{\dfrac{{1000}}{{18}} + 2.5}} = \dfrac{{1000}}{{1045}} $
Now, substituting the values, we will calculate the vapour pressure of $ 2.50{\text{ }}molal $ $ {C_6}{H_{12}}{O_6} $ as stated below:
$ {P_{{C_6}{H_{12}}{O_6}}} = \dfrac{{1000}}{{1045}} \times 23.8 = 22.775{\text{ }}mmHg $
Hence, the vapour pressure of $ 2.50{\text{ }}molal $ $ {C_6}{H_{12}}{O_6} $ is $ 22.775{\text{ }}mmHg $ .
Note :
Molality, a measure of concentration, is especially used when we have to study the characteristics of solutions that are related to vapor pressure as well as alterations in temperature. In such cases, molality is preferred because its value does not vary with alteration in temperature. As in other measures of concentration, volume of solution is used which depends slightly upon temperature.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

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

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

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
