What is the expression for calculating the molar mass of a non-volatile solute from the vapour pressure of the solution?

Answer

Text Solution

Solution : `Ps=80%" of "P^(@)=0.80P^(@),MB=40g//mol` <br> `W_(A)=114g" "M_(A)=114g//mol" "n_(A)=(114)/(114)=1` <br> `(P^(@)-Ps)/(P^(@))=chi_(B)rArr(P^(@)-0.80P^(@))/(P^(@))=((WB)/(40))/((WB)/(4)+1)` <br> `0.2((WB)/(40)+1)=(WB)/(40)rArrWB=10g`

Text Solution

Solution : `(P^(@)-P_(S))/(P_(S)) = (n)/(N) = (wM)/(m xx W)`, Ocatne is `C_(8)H_(18)`. <br> (Given `m = 40, W = 114g, M_("Ocatne") = 114`) <br> `(100-80)/(80) = (w xx 114)/(40 xx 114)` <br> `w = 10 g` <br> Note: By `(P^(@)-P_(S))/(P_(S)) = (n)/(N)` <br> (Only for dilute solution, the answer comes `8 g`.)

1.00 g of nonvolatile sulfanilamide, C6H8O2N2S, is dissolved in 10.0 g of acetone, C3H6O. The vapor pressure of pure acetone at the same temperature is 400 mm Hg.

Calculate the vapor pressure of the solution.

1. What is the question asking you to do?

   Calculate the vapor pressure of the solution.

   Pa = ? mm Hg

2. What information have you been given?

   mass(solute) = mass(C6H8O2N2S(s)) = 1.00 g

   mass(solvent) = mass(C3H6O) = 10.0 g

   vapor pressure of pure solvent = Pao = 400 mm Hg

3. What is the relationship between what you know and what you need to find out?

   Pa = XaPao

   Pao = 400 mm Hg

   Xa = mole fraction of solvent
       = moles(solvent) ÷ (moles(solute) + moles(solvent))

4. Perform the calculations:

   • Calculate moles of solute: n(C6H8O2N2S)
     n(C6H8O2N2S) = mass(C6H8O2N2S) ÷ molar mass(C6H8O2N2S)
         mass(C6H8O2N2S) = 1.00 g
         molar mass(C6H8O2N2S) = (6 x 12.01) + (8 x 1.008) + (2 x 16.00) + (2 x 14.01) + 32.06 = 172.204 g mol-1

     moles(C6H8O2N2S) = 1.00 g ÷ 172.204 = 0.005807 mol

   • Calculate moles of solvent: n(C3H6O)
     n(C3H6O) = mass(C3H6O) ÷ molar mass(C3H6O)
         mass(C3H6O) = 10.0 g
         molar mass(C3H6O) = (3 x 12.01) + (6 x 1.008) + 16.00 = 58.078 g mol-1

     moles(C3H6O) = 10.0 g ÷ 58.078 = 0.1722 mol

   • Calculate the mole fraction of the solvent: Xsolvent
     Xsolvent = molessolvent ÷ (molessolute + molessolvent)
     Xa = moles(C3H6O) ÷ [moles(C3H6O) + moles(C6H8O2N2S)]
         = 0.1722 ÷ [0.1722 + 0.005807] = 0.9674
   • Calculate the vapor pressure: Pa
     Pa = XaPao
     Pa = 0.9674 x 400 mm Hg = 386.96 mm Hg = 387 mm Hg

5. Is your answer plausible?

   Addition of solute to solvent lowers the vapor pressure, that is,
   Pa < Pao
   where Pa is vapor pressure of solution and Pao is vapor pressure of pure solvent
   Pao = 400 mm Hg (given in question)
   Pa = 387 mm Hg (calculated)
   387 < 400 so our answer is plausible.

6. State your solution to the problem.

   Vapor pressure of solution is 387 mm Hg

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Pankaj 2019-05-24 11:24:16

Give the complete & correct solution, immediately

Tanya 2019-05-04 12:11:11

This solution is quite easy but this is not correct. The correct answer for this question is 10.

Tanya 2019-05-04 12:09:55

This is quite easy but correct answer is 10

Bhagya 2019-04-25 10:54:30

For the answer to be 10 , it should be equated with X of solute i.e. ((w1/M1)/(w1/M1)+(w2/M2))

Gautam 2019-04-22 22:21:37

Correct is 10

Neha 2019-03-28 23:44:30

But answer is 10

Neha 2019-03-28 23:43:03

Give the correct solution

Renuka 2019-03-10 23:58:19

Answer is right

Malavika Manish 2019-03-09 15:07:11

The answer is wrong ....the correct answer should be done like given below if vapour pressure of pure liquid is = Po 80 % of pure liquid Ps= 80×Po/100 = 0.8Po Ps =Po × Xsolute mass of solute = x gram mass of solvent = 114g Molar mass of solute= 40 g/mol Molar mass of solvent (octane C8H18) = 114g/mol Number of moles of solute = x/40 = 0.025x Number of moles of solvent = 114/114= 1 moles Mole fraction of solvent = 1/(1+0.025x) 0.8Po=Po×1/(1+0.025x) Cross multiply we get (1+0.025x))0.8Po= Po Divide by 0.8 Po we get 1+0.025x = 1.25 Subtract 1 both side we get 0.025x = 0.25 Now divide by 0.025 we get x = 10g

nikhil gupta 2019-01-17 15:37:35

this answer in wrong instead the formula used will be p-ps/ps

Page 2

Answer

Let, the molar mass of the solute be M g mol - 1

Now, the no. of moles of solvent (water),n1 = 90g / 18g mol-1

And, the no. of moles of solute,n2 = 30g / M mol-1 = 30 / M mol

p1 = 2.8 kPa

Applying the relation:

(p10 - p1) / p10    =  n2 / (n1  + n2)

⇒ (p10 - 2.8) / p10    =  (30/M)  / {5  + (30/M)}

⇒ 1 - (2.8/p10)  = (30/M) / {(5M+30)/M}

⇒ 1 - (2.8/p10) = 30 / (5M + 30)

⇒ 2.8/p10  = 1 -  30 / (5M + 30)

⇒ 2.8/p10   =  (5M + 30 - 30) / (5M + 30)

⇒ 2.8/p10   = 5M / (5M+30)

⇒ p10 / 2.8  =  (5M+30) / 5M  ----------------(1)

After the addition of 18 g of water:

n1 = (90+18g) / 18  = 6 mol

and the new vapour pressure is p1 = 2.9 kPa   (Given)

Again, applying the relation:

(p10 - p1) / p10    =  n2 / (n1  + n2)

⇒ (p10 - 2.9) / p10    =  (30/M)  / {6  + (30/M)}

⇒ 1 - (2.9/p10)  = (30/M) / {(6M+30)/M}

⇒ 1 - (2.9/p10) = 30 / (6M + 30)

⇒ 2.9/p10  = 1 -  30 / (6M + 30)

⇒ 2.9/p10   =  (6M + 30 - 30) / (6M + 30)

⇒ 2.9/p10   = 6M / (6M+30)

⇒ p10 / 2.9  =  (6M+30) / 6M  ----------------(2)

Dividing equation (1) by (2),we get:

2.9 / 2.8 =   {(5M+30) / 5M} / {(6M+30) / 6M}

⇒ 2.9 x (6M+30 / 6)  =  (5M+30 / 5) x 2.8

⇒ 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6

⇒ 87M + 435  =  84M + 504

⇒ 3M = 69

⇒ M = 23u

Therefore, the molar mass of the solute is 23 g mol - 1.

(ii) Putting the value of 'M' in equation (i), we get:

⇒ p10 / 2.8  =  (5M+30) / 5M 

⇒ p10 / 2.8  =  (5x23+30) / 5x23

⇒  p10  =  (145 x 2.8) / 115

⇒  p10  =  3.53

Hence, the vapour pressure of water at 298 K is 3.53 kPa.

Why 2019-07-26 16:11:20

Thank you so much..it really helped me a lot

Meraj alam 2019-07-03 16:21:02

Why are you adding 1 in p°-p/p° ??

Best explaination 2019-05-23 15:42:29

Nice

akash 2019-03-25 18:20:34

the ways u define or tell really good .. rock on pleaze

Arsh 2019-02-19 22:14:50

Really good

Page 3

Twinkle 2020-07-17 18:47:48

Taugh but it makes it easy

Vivek 2019-11-27 11:24:58

Nice answer

Krypton 2019-09-09 13:01:10

Perfect answer

Abhinav priye 2019-04-06 14:09:29

Very good 👍👍👍👍👌👌👌 Keep it up

Allin 2018-11-20 14:55:45

Kf for water and sugar cane will be same for kf of water and glucose

Soumyanetra Basak 2018-07-27 21:32:07

Good way

Prince Godwami 2018-06-10 12:13:31

Perfect answer.

Vishal singh 2018-06-05 21:12:53

Awesome

Page 4

Baba Kareem Das 2018-03-11 20:23:28

I Like this solutions

Satvir singh 2017-05-12 13:20:52

In this question we are calculating the molecular mass of AB2 and AB4 molecule separately,therefore in case molecular mass of AB2 is taken and in other case molecular mass of AB4 is taken

Pinky 2017-05-11 11:28:54

in molality we take the molecular mass of solution in denominator but here there is only molecular mass of compound AB2.Why?

Page 5

Nevil 2019-08-24 14:40:33

Nyc

Uzair shaikh 2019-06-09 15:29:16

C1=mole/volume=weight/molecular weight × 1/volume = 36/180 × 1/1

manisha 2019-05-28 13:09:30

what is the meaning of writing C1= 36/180 but its real value is 36g

Khushi 2019-05-15 22:23:19

How can write 36/180

Obd 2019-05-14 21:29:20

Sir its 0.061

RAUSHAN KUMAR 2019-04-10 20:42:28

How write 36/180

Chandan Sharma 2018-09-17 09:11:30

Sorry... The answer of this question is... 0.06 not 0.006

Page 6

2019-05-12 20:32:12

Thanks

Page 7

Answer

n-octane is an organic solvent (liquid)

Out of given examples cyclohexane is strongest organic solvent, so according to like dissolves like, cyclohexane is most likely to be dissolved as a solute in n octane. After cyclohexane, CH3CN

Will be dissolved completely as a solute in n octane, then CH3OH & the least soluble will be Kcl. Also since n-octane & cyclohexane, both of them belongs to alkane category, their solubility will be maximum.

Therefore the order of increasing solubility in n octane is as follows:

Kcl < CH3OH < CH3CN < Cyclohexane

Candygirl 2019-07-04 20:44:21

What is right??

Chandan Sharma 2018-09-17 09:08:40

The answer of this question (q. N. 23) is wrong..... Because CH3OH is less than CH3CN.... So order Will be like this... Cyclohexan> CH3OH>CH3CN>KCl

Page 8

Farhan khan 2019-09-28 00:39:14

Explain briefly

Page 9

Shyam Sunder 2019-03-27 18:54:52

why density is mentioned?

Page 10

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Question 27

Answer

Solubility product of CuS, Ksp of CuS = 6 x 10-16

If s is the solubility,

then CuS = cu2+ + S2-

Therefore KSP = { cu2+}{ S2-}

Or

KSP = s x s

Or

s = √ KSP = √6 x 10-16

= 2.45 x 10-8 M

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