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1.

When `250 mg` of eugenol is added to `100 g` of camphor `(k_(f)=39.7" molality"^(-1))`, it lowered the freezing point by `0.62^(@)C` . The molar of eugenol is `:`A. `1.6xx10^(2) g//mol`B. `1.6xx10^(4)g//mol`C. `1.6xx10^(3)g//mol`D. `200g//mol`

Answer» Correct Answer - 1
`0.62=(250xx10^(-3))/(Mxx100)xx1000xx39.7xx1`
`M=160` or `1.6xx10^(2)g//mol`
Discussion
2.

Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water?A. `6%` by mass of urea in aqueous solutionB. `100g` of aqueous solution containing `18g` of glucoseC. `59g` of aqueous solution containing `9g` of glucoseD. `1M KCl` solution in water.

Answer» Correct Answer - 3
Cryoscopic constant `K_(f)=DeltaT_(f)` of solution having until molality of normal solutes
Molality of glucose solution in `(3)=(9xx1000)/((59-9)xx180)=1`
Discussion
3.

Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water ?A. `6%` by mass of urea is aqueous solutionB. `100g` of aqueous solution containing `18g` of glucoseC. `1M KCl` solution in waterD. `59g` of aqueous solution containing `9g` of glucose

Answer» Correct Answer - 4
Cryosopic constant `K_(r)+Delta T_(f)` of solution having until molality of normal solutes
Molality of glucose solution in `(3)=(9xx1000)/((59-9)xx180)=1`
Discussion
4.

A `5%` solution of cane sugar (molecular weight =`342`) is isotonic with a `1%` solution of substance `X`. The molecular weight of `X` isA. `34.2`B. `171.2`C. `68.4`D. `136.8`

Answer» Correct Answer - C
`pi_(1)=pi_(2),(5)/(342)xx(1000)/(100)=(1)/(X)xx(1000)/(100)" "rArr" "X=68.4`
Discussion
5.

An aqueous solution of a solute AB has b.p. of `101.08^(@)C( AB` is `100%` ionised at boiling point of the solution ) and freezes at `-1.80 ^(@)C`. Hence, `AB(K_(b)//K_(f)=0.3)`A. is `100%` ionised at the `f.p.` of the solutionB. behaves as non`-` electrolyte at the `f.p.` of the solutionC. forms dimerD. none of the above

Answer» Correct Answer - 2
Given `DeltaT_(b)=1.08^(@)C," "I=2` at boiling pt. of solution.
and `DeltaT_(f)=1.80^(@)C,` and `(k_(b))/(k_(f))=0.3`
so`" "(DeltaT_(b))/(DeltaT_(f))=(I_(b)k_(b)m)/(I_(f)k_(f)m)`
so`" "I_(f)=1`
`i.e.,AB` behaves as non-electrolyte at the f.p. of the solution.
Discussion
6.

`106.2g` 1 molal aqueous solutio of ethylene glycol is cooled to `-3.72^(@)C`. Mass of of ice separated during cooling is `(K_(f0` water `=1.86` freezing point of water `=0^(@)C )`A. `25g`B. `50g`C. `60g`D. `40g`

Answer» Correct Answer - 2
Mass of solute in `1kg` water `=62g`
`:.` Mass of solute in `106.2g` solution `=(62)/(1062)xx106.2=6.2g`
`3.72=1.86xx(6.2xx1000)/(62xxw)`
`w=50" ":.` Mass of ixe separated `=50g`
Discussion
7.

At `25^(@)C`, a solution containing `0.2g` of polyisobutylene in `100mL` of benzene develpoed a rise of `2.4mm` at osmotic equilibrium. Calculate the molecular weight of polyisobutylene if the density of solution is `0.88 g// mL`A. `2.39 xx10^(5)g`B. `33.9xx10^(5)g`C. `43.8xx10^(5)g`D. `78.6xx10^(5)g`

Answer» Correct Answer - 1
Height developed `=2.4mm`
Osmotic pressure `=h.d.g.=(2.4)/(10)xx0.88xx981=207.187`dyne`cm^(-2)`
Now `piv=nRTrArr207.187xx100=(0.2)/(m)xx8.314xx10^&(7)xx298" "(R` in erg,V in mL, using CGS system)
`m=2.39xx10^(5)`
Discussion
8.

An aqueous solution containing `21.6 mg` at a solute in `100 ml` of solution has an osmotic pressure of `3.70 mm` of `Hg` at `25^(@)C`. The molecular wt of solute in `g//mol` isA. 1085B. 9035C. 1355D. 700

Answer» Correct Answer - 1
`piV=(w)/(M)RT`
`(3.70)/(760)xx(100)/(1000)=(21.6xx10^(-3))/(m)xx0.0821xx298`
`m(` molecular mass `)=1085g//mol`
Discussion
9.

The Vapour pressure of solution containing `2g` of `NaCl` in `100g` of water at `100^(@)C` is `753.2mm` of `Hg`, then degree of dissociation of `NaCl`.A. `0.6`B. `0.7`C. `0.8`D. `0.9`

Answer» Correct Answer - 1
`(P^(@)-P_(S))/(P_(S))=(wxxM)/(mxxW)`
`(760-753.2)/(760)=(2xx18)/(mxx100)`
`m_(exp)=36.56`
`{:(NaCl,rarr,Na^(+),+,Cl^(-)),(1-alpha,,alpha,,alpha):}`
`M_(the)/M_(exp)=1+alpha" "rArr" "58.5/36.56=1+alpha`
Discussion
10.

What weight of glucose dissolved in `100g` of water will produce the same lowering of vapour pressure as one gram of urea dissolved in `50g` of water at the same temperatureA. `3g`B. `5g`C. `6g`D. `4g`

Answer» Correct Answer - 3
`(W_(A))/(180)xx(18)/(100)=(1)/(60)xx(18)/(50)`
`W_(A)=6 g`
Discussion
11.

What is the mole ratio of benzene `(P_(B)^(@)=150 t o r r)` and toluence `(P_(tau)^(@)=50 t o rr)` in vapour phase if the given solution has a vapour phase if the given solution has a vapour pressure of `120` torr ?A. `7:1`B. `7:3`C. `8:1`D. `7:8`

Answer» Correct Answer - 1
`P=P_(B)^(@)X_(B)+P_(T)^(@)X_(T)`
`120=150(X_(B))+50(1-X_(B))`
`100" "X_(B)+70`
`X_(B)=0.7`
`Y_(B)=(X_(B)p_(B)^(@))/(P)=(0.7xx150)/(120)=0.075" "(Y_(B))/(Y_(T))=(7)/(1)" "Y_(T)=1-0.875=0.125`
Discussion
12.

At `25^(@)C`, the vapour pressure of pure liquid `A ( mol wt. =40)` is `100 t o r r` , while that of pure liquid `B` is 40 torr, `( mol. Wt. =80 )`. The vapour pressure at `25^(@)C` of a solution containing `20g` of each A and B is :A. 80 torrB. `59.8 t o rr`C. `68 t o rr`D. 48 torr

Answer» Correct Answer - 1
Moles of A `=(20)/(40)rArr0.5,`
Moles of B `=(20)/(80)rArr0.25,`
`X_(A)=(0.5)/(0.5+0.25)=0.67,X_(B)=0.33`
`P_("total")=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)`
`rArr" "100xx0.67+40xx0.33rArr80`torr
Discussion
13.

Mixture of volatile components A and B has total pressure ( in Torr ) `p=265-130x_(A),` where `X_(A)` is mole fraction of `A` in mixture . Hence `P_(A)^(@)+P_(B)^(@)=` ( in T o r r).A. 265B. 135C. 400D. 150

Answer» Correct Answer - C
`P=X_(A)P_(A)^(@)+X_(B)P_(B)^(@)=(P_(A)^(@)-P_(B)^(@))X_(A)+P_(B)^(@)`
So`" "P_(B)^(@)=265`
`P_(A)^(@)-P_(B)^(@)=-130" "P_(A)^(@)=135`
`P_(A)^(@)+P_(B)^(@)=400`
Discussion
14.

The plots of `(1)/(X_(A))` ( on `y-` axis ) vs `(1)/(Y_(A))` ( on `x-` axis) ( where `X_(A)` and `Y_(A)` are the mole fractions of liquid `A` in liquid and vapour phase respectively ) is linear with slpe and `y-` intercept respectively.A. `(P_(A)^(@))/(P_(B)^(@))` and `((P_(A)^(@)-P_(B)^(@)))/(P_(B)^(@))`B. `(P_(B)^(@))/(P_(A)^(@))` and `((P_(A)^(@)-P_(B)^(@)))/(P_(B)^(@))`C. `(P_(A)^(@))/(P_(B)^(@))` and `((P_(B)^(@)-P_(A)^(@)))/(P_(B)^(@))`D. `(P_(B)^(@))/(P_(A)^(@))` and `((P_(B)^(@)-P_(A)^(@)))/(P_(B)^(@))`

Answer» Correct Answer - 3
`P+P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))` and
`P_(A)^(@)X_(A)=Y_(A)P=Y_(A)[P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))]`
so, so, `(1)/(Y_(A))=1+(P_(B)^(@))/(P_A^(@))((1)/(X_(A))-1)" "so, " "x=1+(P_(B)^(@))/(P_(A)^(@))(y-1)`
Hence `rArr(x-1)(P_(B)^(@))/(P_(A)^(@))+1=y" "so," y=mx+C` give the result
Discussion