2031 + Interview Questions in SOLUTIONS IN GENERAL AWARENESS Page 39 InterviewSolution

1901.	At a particular temperature, the vapour pressure of two liquids A and B are 120 and 180 mm of Hg respectively. Two moles of A and 3 Moles of B are mixed to form an ideal solution. What is the vapour pressure for the solution?
Answer» P°_A = 120mm of Hg P°_B = 180 mm of Hg X_A = 2/5 X_B = 3/5 P_A = P°_A x X_A = 120 × 2/5 = 48 mm of Hg P_B = P°_B x X_B = 180 × 3/5 = 108 mm of Hg P_S = P_A + P_B = 108 + 48 = 156 mm of Hg

1901.

At a particular temperature, the vapour pressure of two liquids A and B are 120 and 180 mm of Hg respectively. Two moles of A and 3 Moles of B are mixed to form an ideal solution. What is the vapour pressure for the solution?

Answer»

P°_A = 120mm of Hg

P°_B = 180 mm of Hg

X_A = 2/5

X_B = 3/5

P_A = P°_A x X_A

= 120 × 2/5

= 48 mm of Hg

P_B = P°_B x X_B

= 180 × 3/5

= 108 mm of Hg

P_S = P_A + P_B

= 108 + 48

= 156 mm of Hg

Discussion

1902.	The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
Answer» P°H₂O = 12.3 kPa 1000 In 1 molal solution, n_solute = 1; nH₂O = \(\frac{1000}{8}\) = 55.5 ∴ XH₂O = \(\frac{55.5}{55.5\,+\,1}\) Vapour pressure of the solution, P_s = P^°H₂O × XH₂O = 0.982 × 12.3 = 12.08 kPa

1902.

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Answer»

P°H₂O = 12.3 kPa

1000

In 1 molal solution, n_solute = 1; nH₂O = \(\frac{1000}{8}\) = 55.5

∴ XH₂O = \(\frac{55.5}{55.5\,+\,1}\)

Vapour pressure of the solution, P_s = P^°H₂O × XH₂O

= 0.982 × 12.3 = 12.08 kPa

Discussion

1903.	Mole fraction of `C_(3)H_(5)(OH)_(3)` in a solution of `36g` of water and `46g` of glycerine is:A. `0.46`B. `0.36`C. `0.20`D. `0.40`
Answer» Correct Answer - C Mole of `H_(2)O=(36)/(18)=2` Mole of glycerine `=(46)/(92)=0.5` total mole `=2+0.5=2.5` Mole fractions of glycerine `=(n_(1))/(n_(1)+n_(2))=(0.5)/(2.5)rArr X_(0)=0.2`

Discussion

1904.	The boiling point of pure acetone is 56.38^(@)C`. When 0.707 g of a compound is dissolved in 10 g of acetone there is elevation to `56.88^(@)C` in b.p What is the mol.wt. of the comound ? `(K_(b)" of acetone"=1.72 K kg mole^(-1))`
Answer» `DeltaT_(b)=K_(b)xxm` `rArr (56.88-56.38)=1.72xx(0.707)/(M_(B))cc(100)/(10)` `:. M_(B)=243.2`

Discussion

1905.	The amount of solute (molar mass 60 g mol-1) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
Answer» The amount of solute (molar mass 60 g mol^-1) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is 60 g.

Discussion

1906.	Calculate mole fraction of ethyl alcohol and water in a solution containing 46 g ethyl and 36g water.
Answer» `n_(C_(2)H_(5)OH)=(46)/(46)=1` `n_(H_(2)O)=(36)/(18)=2` `X_(C_(2)H_(5)OH)=(n_(C_(2)H_(5)OH))/(n_(C_(2)H_(5)OH))=(1)/(1+2)=(1)/(3)` `X_(H_(2)O)=1-(1)/(3)=(2)/(3)`

Discussion

1907.	What weight of solute (mol. Wt. 60) is required to dissolve in 180 g of water to reduce the vapour pressure to `4//5^(th)` of pure water ?
Answer» `p^(@)-(p_(s))/(p^(@))=(wxxM)/(MxxW)` `(p^(@)-(4)/(5)p^(@))/(p^(@))=(wxx18)/(60xx180)`

Discussion

1908.	The lowering of vapour pressure in a saturated aq. Solution of salt AB is found to be `0.108` torr. If vapour pressure of pure solvent at the same temperature is 300 torr, find the solubility product of salt AB.A. `10^(-8)`B. `10^(-6)`C. `10^(-4)`D. `10^(-5)`
Answer» Correct Answer - C

Discussion

1909.	Which compound corresponds vant Hoff factor (i) to be equal to 2 in dilute solution:A. `K_(2)SO_(4)`B. `NaHSO_(4)`C. SugarD. `MgSO_(4)`
Answer» Correct Answer - D

Discussion

1910.	`50 mL` of `10 N H_(2)SO_(4), 25 mL` of `12 N HCI` and `40 mL` of `5N HNO_(3)` are mixed and the volume of the mixture is made 1000 mL by adding water. The normality of resulting solution will beA. `1N`B. `2N`C. `3N`D. `4N`
Answer» Correct Answer - A `N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(4)V_(2)` `10 N xx 50 mL +12 Nxx25 mL+5 Nxx40 mL` `=N_(4)xx1000 mL` (Volume of mixture = 1000 mL) `N_(4) = (500+300+200)/(1000) N = 1N`

Discussion

1911.	The vapour pressure of water at `23^(@)C` is 19.8 mm of Hg 0.1 mol of glucose is dissolved in `178.2g` of water. What is the vapour pressure (in mm Hg) of the resultant solution?A. `19.0`B. `19.602`C. `19.402`D. `19.202`
Answer» Correct Answer - B n(glucose) = 0.1 n(water) `= (178.2)/(18) = 9.9` x(water) `=(9.9)/(9.9+0.1)=0.99` p(solution) `= p^(@)` (solvent) `xx` x(solvent) `= (19.8 mm Hg)xx 0.99` `= 19.602 mm Hg`

Discussion

1912.	Ethylene glycol `("molar mass"=62 g mol^(-1))` is a common automobile antyfreeze. Calculate the freezing point of a solution containing 12.4 g of this substance in 100 g of water. (Given` K_(f) for water = 1.86 K kg mol^(-1))`
Answer» `M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))` Mass of ethylene glycol`(W_(B))=12.4 g` Mass of solvent i.e., water `(W_(A))=100g=0.1 kg` Molar mass of glycol `(M_(B))=62 g mol^(-1)` Molal depression constant `(K_(f))=1.86 K kg mol^(-1))` `DeltaT_(f)=((12.4g)xx(1.86 K kgmol^(-1)))/((62 g mol^(-1))xx(0.1 kg))=3.72 K` Freezing point of the solution =(273.0K-3.72 K)= 269.28 K.

Discussion

1913.	When `3`g of a novolatile solute is dissolved in 50 g of water , the relative lowering of vapour pressure observed is `0.018 Nm^(-2)`. Molecular weight of the substances isA. `60`B. `30`C. `40`D. `120`
Answer» Correct Answer - A `(P_(A)^(@)-P)/(P_(A)^(@))=chi_(B),` By putting values `0.018=((w_(B))/(M_(B)))/(w_(A)/(w_(B))+(w_(B))/(M_(B)))` `rArr 0.018=((3)/(M_(B)))/((3)/(M_(B))+(50)/(18))` by calculating `M_(B)~~60`

Discussion

1914.	The vapour pressure of a pure liquid at `25^(@)C` is 100 mm Hg. Calculate the relative lowering of vapour pressure if the mole fraction of solvent in solution is 0.8.
Answer» `p^(@)-p_(s))/(p^(@))=x_("solute")` =0.2

Discussion

1915.	At room temperature the mole fraction of a solution is 0.25 and the vapour pressure of the solvent is 0.80 atm. Then the lowering of vapour pressure isA. 0.75B. 0.6C. 0.2D. 0.8
Answer» Correct Answer - C `(P^(@)-P_(s))/(P^(@)) = x_(2) :. (P^(@)-P_(s))/(0.80) = 0.25` or `P^(@)-P_(s) = 0.20` atm

Discussion

1916.	`45 g` of ethylene glycol `C_(2)H_(6)O_(2)` is mixed with `600 g` of water. Calculate (a) the freezing point depression and (b) the freezing point of solution. Given`K_(f)=1.86 K kg mol^(-1)`.
Answer» Depression in freezing point is realed to the molality, therefore the molality of the solution with respect to ethylene glycol, `DeltaT_(f)=K_(f)m` Mole of ethylene glycol = `45 gxx(1 mol)/(62g) =0.73 mol` Mass of water in kg =`600 xx (1 kg)/(1000 g)=0.60 kg` Hence, molality of ethylene glycol `=(0.73 mol)/(0.60 kg)=1.20 mol kg` Therefore,freezing point depression `(DeltaT_(f)=1.86 K kg mol^(-1))xx(1.2 mol kg^(-1) )=2.2 K` Freezing point of the aqueous solution `T_(f) =T_(f)^(@) - Delta_(f)T` `273.15 K -2.2 K =270.95 K`

Discussion

1917.	Equal volumes of ethylene glycol (molar mass = 62) and water (molar mass = 18) are mixed. The depression in freezing point of water is (given `K_(f)` of water `= 1.86 K mol^(-1)` kg and specific gravity of ethylene glycol is 1.11)A. `0.0033`B. `3.33C. `0.333`D. `33.3`
Answer» `M=DxxV` M (Ethylene glycol) `= 1.11 xx upsilon` M (water) `= 1.00 xx upsilon` n(Ethylene glycol) = `(1.11 upsilon)/(62)` mol `Delta T_(f)= K_(f)xx m=1.86 xx (1.11 upsilon)/(62)xx(1000)/(1.00 upsilon)` `=(1.86xx1.11xx1000)/(62)=33.3`

Discussion

1918.	Relative lowering of vapour pressure , osmotic pressure of a solution and elevation in boiling points are __(p)___ properties. Osmosis is the passage of __(q)__ through a semipermeable membrane from a solution of __(r)__ towards a solution of __(s)__ . Osmotic pressure is equivalent to mechanical pressure which must be applied on __(t)__ to prevent osmosis. In the above paragraph p,q,r,s and t respectively areA. colligative, solution, higher concentration , lower concentration , solutionB. colligative, solvent, higher concentration , lower concentration , solutionC. colligative , solution , lower concentration, higher concentration , solventD. colligative , solvent , lower concentration, higher concentration , solution.
Answer» Correct Answer - D

Discussion

1919.	`12g` of a nonvolatile solute dissolved in `108g` of water produces the relative lowering of vapour pressure of `0.1`. The molecular mass of the solute isA. 60B. 80C. 40D. 20
Answer» Correct Answer - D

Discussion

1920.	Statement-I: Relative lowering of vapour pressure is equal to mole fraction of the solvent. Because Statement-II: Relative lowering of vapour pressure is a colligative property.
Answer» Correct Answer - D

Discussion

1921.	45 g of ethylene glycol `(C_(2)H_(6)O_(2))` is mixed with 650 g of water. The freezing point depression (in K) will be `(K_(f)"for water "=1.86 "K kg mol"^(-1))`.
Answer» Molality of solution (m) `=("No of moles of "C_(2)H_(6)O_(2))/("Mass of water in kg")` `=(((45g))/((62" g mol"^(-1))))/(6.65kg)=((0.73 mol))/((0.65kg))=1.12"mol kg"^(-1)` `DeltaT_(f)=K_(f)xxm=(1.86" K kg mol"^(-1))xx(1.12 "mol kg"^(-1))` `=2.08 K~~2K.`

Discussion

1922.	Lowering in vapour pressure is highest forA. `0.2 m urea`B. `0.1 m glucose`C. `0.1 m MgSO_(4)`D. `0.1 m BaCl_(2)`
Answer» Correct Answer - D `(P^(@)-P_(S))/P^(@)`=Molality `xx (1 - alpha + xalpha +yalpha)` The value of `P^(@)-P_(S)` is maximum for `BaCl_(2)`.

Discussion

1923.	Which of the following will have the highest boiling point at `1 atm` pressure?A. `0.1 M NaCl`B. `0.1 M sucrose`C. `0.1 M BaCl_(2)`D. `0.1 M glucose`
Answer» Correct Answer - C `DeltaT_(b)=K_(b) xx Molality xx (1- alpha + xalpha + yalpha)` `DeltaT_(b)` is more `BaCl_(2)` as it gives more number of ions, `(1- alpha + xalpha + yalpha)` is more for `BaCl_(2)` (it is 3).

Discussion

1924.	Does it advise to use ethylene glycol in radiators of cars in summer?
Answer» No, it is not advised to use ethylene glycol in radiators of cars in summer because ethylene glycol decreases the freezing point of water, prevents the freezing of water of radiator in winter, so it is advised to use ethylene glycol in winter.

Discussion

1925.	`A 0.50` molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point constant of water is `1.86^(@)` per molal, at which temperature will the mixture freeze?
Answer» Correct Answer - B `DeltaT_(f)=k_(f)m`. `DeltaT_(f)=1.86xx0.5=0.93`. so `T_(f)=-0.93^(@)C`.

Discussion

1926.	An azeotropic mixture of `HCl` and water hasA. `84%` of `HCl`B. `22.2%` of `HCl`C. `63%` of `HCl`D. `20.2%` of `HCl`
Answer» Correct Answer - D Azeotropic mixture of `HCl` contains `20.4%HCl`.

Discussion

1927.	35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.
Answer» 35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O. Molar mass of water = 18 g mol^-1 Molar mass of ethylene glycol (CH₂OHCH₂OH) = 62 g mol^-1 Number of moles of water = n₁ = \(\frac{65}{18}\)= 3.611 mol Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)= 0.5645 mol Total moles = n = n₁ + n₂ = 3.611 + 0.5645 = 4.1755 mol Mole fraction of ethylene glycol = x₂ = \(\frac{n_2}{n_1}\)= \(\frac{0.5645}{4.1755}\) = 0.1352 ∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352 = 0.8648 ∴ Mole fraction of water = 0.8648 Mole fraction of ethylene glycol = 0.1352

1927.

35 % (W/W) solution of ethylene glycol in water, an anti-freezer used in automobiles in radiators as a coolant. It lowers freezing point of water to -17.6 °C. Calculate the mole fraction of the components.

Answer»

35% (W/W) means 100 g solution contains 35 g ethylene glycol (CH₂OH-CH₂OH) and 65 g H₂O.

Molar mass of water = 18 g mol^-1

Molar mass of ethylene glycol (CH₂OHCH₂OH) = 62 g mol^-1

Number of moles of water = n₁ = \(\frac{65}{18}\)= 3.611 mol

Number of moles of ethylene glycol = n₂ = \(\frac{35}{62}\)= 0.5645 mol

Total moles = n = n₁ + n₂

= 3.611 + 0.5645

= 4.1755 mol

Mole fraction of ethylene glycol = x₂

= \(\frac{n_2}{n_1}\)= \(\frac{0.5645}{4.1755}\) = 0.1352

∴ Mole fraction of water = 1 – x₂ = 1 – 0.1352

= 0.8648

∴ Mole fraction of water = 0.8648 Mole fraction of ethylene glycol = 0.1352

Discussion

1928.	Pure water boils at 373 K and pure nitric acid boils at 359 K. The azeotropic mixture of water and nitric acid boils at T.K.A. `T lt 359 K`B. `T gt 359 K`C. `T lt 373 K"but"gt 359 K`D. Unpredictable
Answer» Correct Answer - 2

Discussion

1929.	Can we sparete the compnents of azeotropic mixture by fractional destillation ?
Answer» No, it is not possible.

Discussion

1930.	An azeotropic mixture of `HCl` and water hasA. `48% HCI`B. `22.2% HCI`C. `36% HCI`D. `20.2% HCI`
Answer» Correct Answer - D Azeotropic mixture of HCl and water contains `20.24 %` of HCl. It has b.pt. of `108.5^(@)C` under normal pressure.

Discussion

1931.	The vapour pressure of ethanol and methanol are `44.0 mm Hg` and `88.0 mm Hg`, respectively. An ideal solution is formed at the same temperature by mixing `60 g` of ethanol with `40g` of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour.A. `0.66`B. `0.55`C. `0.11`D. `0.33`
Answer» Correct Answer - A Mole of `CH_(3)CH=(40)/(32)=1.25` Mole of `CH_(3)CH_(2)OH=(60)/(46)=1.304` Let `CH_(3)CHrarrA,C_(2)H_(5)OHrarrB` `X_(A)=(n_(A))/(n_(A)+n_(B))=(1.25)/(1.25+1.304)=0.49` `:.X_(B)=0.51` `:.P_(T)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)=88.7xx0.49+44.5xx0.51` `:.P_(T)=43.480+2.695=66.175` `P_(A)=43.8 , :. X_(A)=(P_(A))/(P_(T))=(43.48)/(66.175)=0.66`

Discussion

1932.	STATEMENT-1 : 1M `H_(2)SO_(4)` is also 1 N. STATEMENT-2 : Water glycol mixture is used in car radiators in winter because its freezing point is less than `0^(@)C` STATEMENT-3 : A saturated solution will remain saturated at all temperatures.A. T F TB. F T TC. F T FD. T F F
Answer» Correct Answer - 3

Discussion

1933.	STATEMENT-1 : At definite temperature, the solubility of a solute is fixed. STATEMENT-2 : When azeotropic mixture is distilled, its composition remains same. STATEMENT-3 : Vapour pressure is a colligative property.A. F F TB. T T FC. F T FD. T T T
Answer» Correct Answer - 2

Discussion

1934.	If the total vapour presure of the liquid mixture`A`and`B`is given by the equation:`P=180X_(A)+90`then the ratio of the vapour pressure of the pure liquids A and B is given by :A. `3:2`B. `4:1`C. `3:1`D. `6:2`
Answer» Correct Answer - C As we know `PT=P_(A)^(@)X_(S)+P_(B)^(@)X_(A)` `=P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))=(P_(A)^(@)-P_(B)^(@))X_(A)+P_(B)^(@)…(1)` But from question,`P_(T)=180X_(A)+90...(2)` Equating equation(1)and(2) `:.P_(A)^(@)-P_(B)^(@)=180` `P_(B)^(@)=90,P_(A)^(@)=180+90=270` `(P_(A)^(@))/(P_(B)^(@))=(270)/(90)=3:1`

Discussion

1935.	Effect of adding a non-volatile solute to a solvent is:A. to lower the vapour pressureB. to increase the freezing pointC. to increase the boiling pointD. to decrease the osmotic pressure.
Answer» Correct Answer - a,c (a,c) On adding a non-volatile solute to the solvent, the vapour pressure of the solution gets lowered and its boiling point is reised.

Discussion

1936.	Vapour pressure of `"C"_(6)"H"_(6)" and""C"_(7)"H"_(8)` mixture at `50^(@)"C"` is given by P(mm Hg)`=180X_("B")+90`, where `"X"_("B")` is the mole fraction of `"C"_(6)"H"_(6).` A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapour over this solution ar removed and condensed into liquid and again brought to the temperature of `50^(@)"C"`, what would be the new mole fraction of `"C"_(6)"H"_(6)` in the vapour state ?
Answer» Correct Answer - 0.93 `"C"_(6)"H"_(6)-"B,"" C"_(6)"H"_(7)-"T"` `"n"_("B")=936/78," n"_("T")=736/92=8` `"X"_("B")=("n"_("B"))/("n"_("B")+"n"_("T"))=12/(12+8)=0.6` Now `"P"=("P"_("B")^(0)-"P"_("T")^(0))"X"_("B")+"P"_("T")^(0)` `"P"_("A")^(0)=90"mm"," P"_("T")^(0)=270"mm"` `"P"_("T")=0.6xx270," P"=180xx0.6+90` `=162"mm "=198"mm"` `"Y"_("B")=("P"_("B"))/"P"=162/198=9/11` Mole fraction of benzene in vapour Now on condensation `"X"_("B")="Y"_("B")" X"_("B")=9/11` `"P"_("B")=9/11xx270"mm"," P"=180xx9/11+90=90(29/11)"mm"` `"X"_("B")="P"_("B")/"P"=(9/11xx270)/(90xx29/11)=27/29=0.93`

Discussion

1937.	STATEMENT-1 : Effect of adding a non-volatile solute to a solvent is to increases its freezing point. STATEMENT-2 : Molality is a dimensionless quantity. STATEMENT-3 : The hard shell of an egg was dissolved in HCl solution, and then egg was placed in concentrated solution of NaCl. Then egg will shrink.A. T F TB. F F TC. T T FD. F T F
Answer» Correct Answer - 2

Discussion

1938.	STATEMENT-1 : For solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. STATEMENT-2 : Always there will be lowering of vapour pressure on addition of non-volatile solute to a solvent. STATEMENT-3 : If there is dissociation of non-volatile then the V.P. of solution increases.A. T F TB. F F TC. T T FD. F T F
Answer» Correct Answer - 3

Discussion

1939.	`P_(A)and P_(B)` are the vapour pressure of pure liquid components ,Aand B respectively of an ideal binary solution,If `x_(A)` represents the mole fraction of component A, the total pressure of the solution will beA. `P_(A) + X_(A) (P_(B) - P_(A))`B. `P_(A) + X_(A) (P_(A) - P_(B))`C. `P_(B) +X_(A) (P_(B) - P_(A))`D. `P_(B) +X_(A) (P_(A)-P_(B))`
Answer» Correct Answer - D `P_("Total")=X_(A)P_(A)+X_(B)P_(B)=X_(A)P_(A)+(1-X_(A))P_(B)` `P_(B)+X_(A)(P_(A)-P_(B))`

Discussion

1940.	P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ……(a) P1 + x1(P2 – P1) (b) P2 – x1(P2 + P1) (c) P1 – x2(P1 – P2) (d) P1 + x2(P1 – P2)
Answer» (c) P₁ – x₂ (P₁ – P₂) P_total = P₁ + P₂ = P₁x₁ + P₂x₂ = P₁(1 – x₂ ) + P₂x₂ = P₁ – P₁x₂ + P₂x₂ = P₁ – x₂(P₁ – P₂ ) [∵ x₁ + x₂ = 1 x₁ = 1 – x₂]

1940.

P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ……(a) P1 + x1(P2 – P1) (b) P2 – x1(P2 + P1) (c) P1 – x2(P1 – P2) (d) P1 + x2(P1 – P2)

Answer»

(c) P₁ – x₂ (P₁ – P₂)

P_total = P₁ + P₂

= P₁x₁ + P₂x₂

= P₁(1 – x₂ ) + P₂x₂

= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂(P₁ – P₂ )

[∵ x₁ + x₂ = 1

x₁ = 1 – x₂]

Discussion

1941.	When an ideal binary solution is in equilibrium with its vapour, molar ratio of the two components in the solution and in the vapur phase is :A. same boiling point but different freezing pointB. DifferentC. May or may not be same depending upon volatile nature of the two componentsD. None of the above
Answer» Correct Answer - C Volalite nature `uparrow`. Vapour pressure increases, fraction in vapour phase increases.

Discussion

1942.	Statement-1:All colloidal dispersions give very low osmotic pressure and show very small freezing point depression or boiling point elevation. Statement-2:Tyndall effect is due to scattering of light from the surface of colloidal particles.A. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1B. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-2C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true
Answer» Correct Answer - B Colloidal particles have high molar mass, so their mole fraction is very less causing low colligative properties.

Discussion

1943.	Assertion: The Brownian movement is due to the bombardment on colloidal particle by the molecules of dispersion midium which are in the constant motion like molecules in a gas. Reason: Brownian movement provides a visible proof of the random kinetic motion of molecules in a liquid.A. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-2B. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-3C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true
Answer» Correct Answer - B On the basis of Brownian movement, we can say that molecules are in contineous motion.

Discussion

1944.	Arsenic (III) sulphide forms a sol with a negative charge. Which of the following ionic substances should be most effective in coagulating the sol?A. KClB. `MgCl_(2)`C. `Al_(2)(SO_(4))_(3)`D. `Na_(3)PO_(4)`
Answer» Correct Answer - C More the charge on cation, more the effectiveness of the electrolyte.

Discussion

1945.	when `1.80 g` of a nonvolatile solute is dissolved in `90 g` of benzene, the boiling point is raised to `354.11K`. If the boiling point of benzene is `353.23K` and `K_(b)` for benzene is `2.53 KKg mol^(-1)`, calculate the molecular mass of the solute. Strategy: From the boiling point of the solution, calculate the boiling point elevation, `DeltaT_(b)`, then solve the equation `DeltaT_(b)=K_(b)m` for the molality `m`. Molality equals moles of solute divided by kilograms of solvent (benzene). By substituting values for molality and kilograms `C_(6)H_(6)`, we can solve for moles of solute. The molar mass of solute equals mass of solute `(1.80 g)` divided by moles of solute. The molecular mass (in amu) has the same numerical value as molar mass in `g mol^(-1)`.
Answer» Step 1. Calculate the molality The boiling point elevation is `DeltaT_(b)=(T_(b)-T_(b)^(0))=(354.11- 353.23)K` =0.88 K So the molality of the solution is `m=(DeltaT_(b))/K_(b)=(0.88 K)/(2.53 K Kg mol^(-1)) =0.348 mol Kg^(-1)` Step 2. Calculate the moles of solute The molality of an aqueous solution is `Molaity=("moles of solute")/("Kg benzene")` Hence, for given solution `0.348 mol Kg^(-1)=("moles of solute")/(90xx10^(-3)Kg)` Rearranging this equation, we get Moles of solute `=(0.348 mol kg^(-1))(90xx10^(-3) Kg)` `=3.132xx10^(-2) mol` Step 3. Calculate the molar mass of solute Molar mass=`(mass)_(solute)/("moles"_(solute))` =`(1.80 g)/(3.132xx10^(-2) mol)` =`57.5 g mol^(-1)` Thus, the molecular mass is 57.5 amu. Alternatively, we can use Equation (2.65) to get the molar mass `M_(2)(g mol^(-1))=(K_(b)(K Kg mol^(-1))(w_(2) g)(1000 g Kg^(-1)))/(DeltaT_(b)(K)(w_(1)g))` `M_(2)=((2.53 K Kg mol^(-1))(1.8 g)(1000 g Kg^(-1)))/((0.88 K)(90 g))` =`57.5 g mol^(-1)`

Discussion

1946.	18g of glucose, `C_(6)H_(12)O_(6)`, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar ? `K_(b)` for water is 0.52 kg `"mol"^(-1)`.
Answer» Moles of glucose `=18"g"//180" g mol"^(-1)=0.1" mol"` Number of kilograms of solvent `=1" kg"` Thus molality of glucose solution `=0.1" mol kg"^(-1)` For water, change in boiling point `DeltaT_(b)=K_(b)xx m=0.52" K kg mol"^(-1)xx0.1" mol kg"^(-1)=0.052" K"` Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be `373.15+0.052=373.202" K".`

Discussion

1947.	The boiling point of benzene is `353.23 K`. When `1.80 g` of a non-volatile solute was dissolved in `90 g` benzene, the boiling point is raised to`354.11 K`. Calculate the molar mass of the solute. (`K_(b)` for benzene is `2.53 K kg mol^(-1)`)
Answer» Given values are: `T_(("benzene"))^(@) = 353.00 K , K_(b) = 2.53.00 K kg mol_(-1)` `T_(b("solution")) = 354.00 K` `W_("solute") = 1.80 g` `W_("solvent") = 90 g` The elevation in boiling point ,`DeltaT_(b) = T_(b("solution")) - T_(b("solvent"))^(@)` `= 354.11 - 353.23` `=0.88 K` Molar mass of solute is given as `Mw_("solute") = (K_(b) xx 1000 xx W_("solute"))/(DeltaT_(b) xx W_("solvent")` `Mw_("solute") = (2.53 xx 1000 xx 1.80)/(0.88 xx 90) = 58.0 g mol^(-1)` Hence, the molar mass of solute is `58.0 g mol^(-1)`.

Discussion

1948.	`H_(2)O (1) hArr H_(2) O(g)` at `373K,DeltaH^(@)=8.31kcal mol^(-1)` Thus, boiling point of 0.1 molal sucrose solution isA. `373.52K`B. `373.052K`C. `373.06K`D. `374.52K`
Answer» Correct Answer - C `H_(2)O(l)` changes to steam `H_(2)O (g)`at`373K`.Thus , this represent latent heat of vaporization .`K_(b)`(molal elevaion constant)is related to `DeltaH^(@)` and boiling point by equation, `K_(b)=(RT_(0)^(2))/(1000cancelOH^(@)` Here,`DeltaH^(@)` is in energy unit per gram of solvent. `=(8.31)/(18)kcal^(-1)` `:.K_(b)=(0.002xx(373)^(2))/(1000xx((8.31)/(18)))` `=(278.25)/(461.66)=0.60^(0)mol^(-1)kg` `DeltaT`(surrose solution = molarity `xx K_(b)=0.1xx0.60=0.06^(0)` `:.` Boiling point of solution `= 373+0.06^(@)=373.06K`

Discussion

1949.	When 1.5 g of a non-volatile solute was dissoved in 90 g of benzene, the boiling point of benzene was raised from `353.23^(-1)` K to 353.93 K. Calculate the molar mas of solute (`K_(b)` for benzene=2.52 K kg `mol^(-1)`).
Answer» Correct Answer - 60.0 g `mol^(-1)` `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xx W_(A))` `W_(B)=1.5g , W_(A)=90.0 g=0.09" kg", DeltaT_(b)=0.7" K"` `K_(b)=2.52" k kg mol"^(-1)` `M_(B)=((2.52" K kg mol"^(-1))xx(1.5" g"))/((0.7" K")xx(0.09" kg"))=60.0" g mol"^(-1)`.

Discussion

1950.	What would be the molar mas of a compound if 6.21 g of it dissolved in 24.0 g of chloroform a solution that has a boiling point of `68.04^(@)C`? Given that the boiling point of pure chloroform is `61.7^(@)C` and `K_(b) for chloroform= 3.63^(@)C/m`.
Answer» `M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A))` `K_(b)=3.63^(@)C//m=3.63 K kg mol^(-1),W_(B)=6.21g,W_(A)=24.0g=0.024 kg,` `DeltaT_(b)=(68.04-61.7)^(@)C=6.34K` `M_(B)=((3.63K kg mol^(-1))xx(6.21))/((6.34 K)xx(0.024 kg))=148.15 g mol^(-1)`

Discussion

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