2031 + Interview Questions in SOLUTIONS IN GENERAL AWARENESS Page 41 InterviewSolution

2001.	The temperature at which molarity of pure water is equal to its molality is `:`A. 273 KB. 298 KC. 277 KD. None of these
Answer» Correct Answer - C Molality and mole fraction are related as follows: `m = (x_(B) xx 1000)/((1 - x_(B)) m_(A))` `1 = (x_(B) xx 1000)/((1 - x_(B)) xx 18)` [ `m = 1 x_(B) =` mole fraction of solute `m_(A)=` molar mass of solvent] `x_(B) = 0.0176, " " x_(A) = 0.9824` `p = p_(0) x_(A)` `p = 760 xx 0.9824` `= 746.624` `Delta p = P_(0) - p = 760 - 746.624` `~~ 13.4` mm Hg

Discussion

2002.	Which is better method for expressing concentration of solution – Molarity or Molality
Answer» Molality is a better method for expressing concentration of solution

Discussion

2003.	What are minimum boiling and maximum boiling azeotropes?
Answer» Minimum boiling azeotropes are the non-ideal solutions showing positive deviation while maximum boiling azeotropes are those which show negative deviation. Because of positive deviation their vapour pressures are comparatively higher and so they boil at lower temperatures while in case of negative deviation, the vapour pressures are lesser and so higher temperature are required for boiling them.

Discussion

2004.	`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` barA. 0.94 barB. 9.4 barC. 0.094 barD. `9.4 xx 10^(-4)` bar
Answer» Correct Answer - C `alpha = (i-1)/(n-1)` `0.9=(i-1)/(2-1),i=1.9` `pi=iCRT` `=1.9xx0.002xx0.082xx300` `=0.094` bar]

Discussion

2005.	Define Azeotropes? What are maximum and minimum boiling azeotropes ? Explain with example.
Answer» Azeotropes are constant boiling mixture which has same composition in liquid phase as well as in vapour phase. The non ideal solutions which exhibit negative deviation from ideal solution at a particular composition are called as maximum boiling azeotropes. e.g 68% aqueous solution of HNO₃ or any other suitable example. The non ideal solutions which exhibit positive deviation from ideal solution at a particular composition are called as minimum boiling azeotropes. _e.g 95% aqueous ethanol by volume or any other suitable example.

2005.

Define Azeotropes? What are maximum and minimum boiling azeotropes ? Explain with example.

Answer»

Azeotropes are constant boiling mixture which has same composition in liquid phase as well as in vapour phase.

The non ideal solutions which exhibit negative deviation from ideal solution at a particular composition are called as maximum boiling azeotropes.

e.g 68% aqueous solution of HNO₃ or any other suitable example.

The non ideal solutions which exhibit positive deviation from ideal solution at a particular composition are called as minimum boiling azeotropes.

_e.g 95% aqueous ethanol by volume or any other suitable example.

Discussion

2006.	`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` bar
Answer» c.`alpha=(i-1)/(m-1)` `0.9=(i-1)/(2-1)`,`i-1.9` Alternate method of calculate `(i)` `i=("Number of ions" xxalpha)+(1+alpha)` `=(2xx0.9)+(1-0.9)` [`alpha=90%` or `0.9`] `=1.8+0.1=1.9` `pi=iCRT` `=1.9xx0.002xx0.082xx300` `=0.094` bar

Discussion

2007.	Calculate the amount of KCl which must be added to 1Kg of water so that the freezing point is depressed by 2 kelvin. (Kf for water = 1.86 K Kg mol-1)
Answer» ΔT_f = i X K_f X m 2 = 2 X 1.86 X ( n/1) N = 1/ 1.86 = 0.537 mol

2007.

Calculate the amount of KCl which must be added to 1Kg of water so that the freezing point is depressed by 2 kelvin. (Kf for water = 1.86 K Kg mol-1)

Answer»

ΔT_f = i X K_f X m

2 = 2 X 1.86 X ( n/1)

N = 1/ 1.86

= 0.537 mol

Discussion

2008.	Molarity and molality of a solution of caustic soda are respectively ` 11.12 M ` and ` 94.12 m ` The density of the solution isA. ` 0.556 g mL^(-1) `B. ` 5.56 g mL^(-1) `C. ` 55.6 gmL^(-1) `D. none of these
Answer» Correct Answer - A ` d = M((1)/(m)+("molar mass of solute")/(1000))` `d = 11.12 ((1)/(94.12)+(40)/(1000))` `=0.556 g mL^(-1) `

Discussion

2009.	Molarity and molality of a solution of caustic soda are respectively ` 11.12 M ` and ` 94.12 m ` The density of the solution isA. `0.556 g mL^(-1)`B. `5.56 g mL^(-1)`C. `55.6 g mL^(-1)`D. None of these
Answer» Correct Answer - A same boiling point

Discussion

2010.	Calcualate the amount of `NaCl` which must be added to `100 g` water so that the freezing point, depressed by `2 K`. For water `K_(f)` =`1.86 K kg mol^(-1)`.
Answer» `NaCl` is a strong electrolyte. It is completely dissociated in solution. Degree of dissociation, `alpha=1` `NaClhArrunderset((n=2))(Na^o++Cl^(ө))` Number of particles after dissociation =`1+(n-1)alpha` `=1+(2-1)xx1` `=2` `DetlaT_(obs)/DetlaT_(theo)="Number of particles after dissociation"/"Number of particles when there is no dissociation"` `2/DetlaT_(theo)=2` ot `DetlaT_(theo)=1` Let W g of NaCl be dissolved in `100 g` of water. So `DetlaT_(theo)=(1000xxK_(f)xxW_(2))/(W_(1)xxMw_(1))` or`W_(2)=(DetlaT_(theo)xxW_(1)xxMw_(1))/(1000xxK_(f))` `(1xx100xx58.5)/(1000xx1.86) = 3.145 g`

Discussion

2011.	10.0 grams of caustic soda when dissolved in 250 cm3 of water, the resultant molarity of solution is :(a) 0.25 M (b) 0.5 M(c) 1.0 M (d) 0.1 M
Answer» Option : (c) 1.0 M

Discussion

2012.	4.0g of caustic soda is dissolved in 100 cc of solution. The normality of solution isA. 1B. 0.1C. 0.5D. 4
Answer» Correct Answer - A `N = (4 xx 1000)/(100 xx 40) = 1`

Discussion

2013.	Isotonic solutions have sameA. Molar concentrationB. MolalityC. NormalityD. None of these
Answer» Correct Answer - A Isotonic solutions have same molar concentration.

Discussion

2014.	A solution containing `0.85 g` of `ZnCI_(2)` in `125.0g` of water freezes at `-0.23^(@)C`. The apparent degree of dissociation of the salt is: `(k_(f)` for water `= 1.86 K kg mol^(-1)`, atomic mass, `Zn = 65.3` and `CI = 35.5)`A. `0.875 M`B. `1.00M`C. `1.75M`D. `0.975M`
Answer» Correct Answer - A `M_("resultant") = (M_(1)V_(1)+M_(2)V_(2))/(V_("total"))` `= (0.5 xx 750 + 2xx 250)/(1000)` `= (375 +500)/(1000) = 0.875M`

Discussion

2015.	A `0.01M` solution of glucose in water freezes at `-0.0186^(@)C`. A `0.01M` solution of KCI in water will freeze at temperatureA. higher than `-0.0186^(@)C`B. `0^(@)C`C. `0.0186^(@)C`D. lower than `-0.0186^(@)C`
Answer» Correct Answer - D `DeltaT_(f)` is more in KCI for same molar conc. Because it ionises in aqueous solution.

Discussion

2016.	Normal boiling point of water is `373 K`. Vapour pressure of water at `298 K` is `23 mm` enthalpy of vaporisation is `40.656 kJ mol^(-1)` if atmopheric pressure becomes `23 mm`, the water will boil at:A. `250K`B. `298K`C. `51.6k`D. `12.5K`
Answer» Correct Answer - B Boiling pt is the temperature at which vapour pressure becomes equal to atmospheric pressure.

Discussion

2017.	Normal boiling point of water is `373 K`. Vapour pressure of water at `298 K` is `23 mm` enthalpy of vaporisation is `40.656 kJ mol^(-1)` if atmopheric pressure becomes `23 mm`, the water will boil at:A. `250K`B. `294K`C. `51.6`D. `12.5K`
Answer» Correct Answer - B Applying Clausis-Clapeyron equation `"log"(P_(2))/(P_(1))=(DeltaH_(V))/(2.303R)[(T_(2)-T_(1))/(T_(1)xxT_(2))]` `"log"(760)/(23)=(40656)/(2.303xx8.314)[(373-T_(1))/(373T)]` This given `T_(1)=294.4K`

Discussion

2018.	A `0.01m` aqueous solution of `K_(3)[Fe(CN)_(6)]` freezes ar `-0.062^(@)C`. What is the apparent percentage of dissociation? `[K_(f)` for water `= 1.86]`
Answer» Correct Answer - `0.78%` `{:(DeltaT_(f) = iK_(f)m ,0.062 =ixx 1.86 xx 0.01),(i = 3.33 ,alpha = (i-1)/(n-1)),(alpha rArr (3.33-1)/(4-1) ,alpha rArr 0.777):}`

Discussion

2019.	0.01 molal aqueous solution of `K_(3)[Fe(CN)_(6)]` freezes at `-0.062^(@)C`. Calculate percentage dissociation `(k_(f)=1.86)`
Answer» Correct Answer - 0.78

Discussion

2020.	A `0.2 molal` solution of `KCl` freezes at `-0.68^(@)C`. If `K_(f)` for `H_(2)O` is `1.86`, the degree of dissociation of `KCl` is a.`85%` , b.`83%` , c. `65%` , d. `90%`
Answer» `DeltaT=ixxK_(f)xxm` `0.68=I xx 1.86 xx0.2` `i=1.83` `alpha=(i-1)/(n-1)=(1.83-1)/(2-1)=0.83` `Ionization=83%`

Discussion

2021.	A `0.2 molal` solution of `KCl` freezes at `-0.68^(@)C`. If `K_(f)` for `H_(2)O` is `1.86`, the degree of dissociation of `KCl` is a.`85%` , b.`83%` , c. `65%` , d. `90%`A. 0.75B. 0.83C. 0.65D. 0.92
Answer» Correct Answer - B `DeltaT=ixxK_(f)xxm` `0.68=ixx1.86xx0.2` `i=1.83` `alpha=(i-1)/(n-1)` `=(1.83-1)/(2-1)=0.83` Ionization = 83%].

Discussion

2022.	`6xx10^(-3)` g oxygen is dissolved in per kg of sea water. Calculate the ppm of exgen in sea water.
Answer» Correct Answer - 6 ppm Mass of exygen `(O_(2))=6xx10^(-3)g` Mass of sea water = 1000g `ppm(O_(2))=("Mass of oxygen")/("Mass of sea water")xx10^(6)=((6xx10^(-3)g))/((10^(3)g))xx10^(6)=6ppm`.

Discussion

2023.	A solution is prepared by dissolving 15g of cane sugar in 60g water. Compute the mass per cent of each component of solution.
Answer» Correct Answer - Mass % of sugar=20; Mass of water = 80 `"Mass % of sugar" = ("Mass of sugar")/("Mass of sugar + Mass of water")xx100` `=((15g))/((75g))xx100=20` `"Mass % of water"=100-20=80`.

Discussion

2024.	A solution is prepared by dissolving 5.64g of flucose in 60g of water. Calculate the following : (i) mass percent of each of glucose and water (ii) molality of the solution (iii) mole fraction of each of glucose and water.
Answer» (i) Total mass of solution `=5.64 +60 =65.64g` Mass percent of glucose `= (5.64)/(65.64)xx100 = 8.59%` Mass percent of water = (100 - Mass percent of glucose) `= (100-8.59)=91.41%` (ii) No. of moles of glucose `=(5.64)/(180)` Mass of water in kg `= (60)/(1000)` Molality `= (5.64)/(180)xx(1000)/(60) = 0.522 m` (iii) No. of moles of glucose `= (5.64)/(180)=0.0313` No. of moles of water `= (60)/(18) = 3.333` Mole fraction of glucose `= (0.0313)/(3.333+0.0313) = 0.0093` Mole fraction of water `= (3.333)/(3.333 + 0.0313)=0.9907`.

Discussion

2025.	What is osmolarity of`0.20MKCl` solution?A. 0.10 osmolB. 0.20 osmolC. 0.30 osmolD. 0.40 osmol
Answer» Correct Answer - D `KCl hArr K^(+) + Cl^(-)` osmosis is a colligative property and thus,osmolarity of `0.2M KCL=0.4` osmol.

Discussion

2026.	In the production of quinoline this compound is isolated from the reaction mixture by steam distillation. Calculate (a) at what temperature the mixture of water and quinoline will boil under a pressure of `740,mm Hg`. (b) What is the maximum number of grams of quinoline which can be distilled with `1000g` water vapour under this pressure. The temperature dependence of vapour pressure. The temprature dependence of vapour pressure of water and quionoline is given as follows : Molecular weight of quinoline `=129`
Answer» Correct Answer - (a) 98.54`. (a) This is an immiscible mixture of water and quinoline, so `P_(1)^(@)+P_(2)^(@)=740` By the help of table, we can say `=98.5+((99-98.5))/((741.21727.95))xx(740-727.95)=98.54`.

Discussion

2027.	Which has maximum osmotic pressure at temperature`TK`?A. 150ml of 1.0M urea solutionB. 210ml of 1.0M urea solutionC. Mixture of 150ml of 1.0M urea solution and 210 mL of 1.0M glucose solutionD. Allof the above are isotonic solutions
Answer» Correct Answer - D `pi=MRT` (a)`pi("urea")=Rt` `(b)pi("glucose")=RT` ( C)`M("mixture")=(M_(1)V_(1)+M_(2)V_(2))/(V_(1)+V_(2))=(150xx1+210xx1)/(150+210)=1M` `:.pi("mixture")=RT` Thus, equal osmotic pressure. Thus, isotonic.

Discussion

2028.	Which has the highest boiling point?A. `0.1m N_(2)SO_(4)`B. `0.1m Al(NO_(3))_(3)`C. `0.1m MgCl_(2)`D. `0.1m C_(6)H_(12)O_(6)("glucose")`
Answer» Correct Answer - C `T_(b)("boiling point")=T_(0)("solvent")+DeltaT_(b)` `DeltaT_(b)=mxxK_(b)xxi` m(same)=0.1molal `DeltaT_(b)propi` Greater the value of i, larger the boiling point of solution `{:((a),Na_(2)SO_(4)hArr2Na^(+)+SO_(4)^(2-)",",(y)/(3)(1)/(1+(y-1)x)=(1+2x)),((b),C_(6)H_(12)O_(6)"(non-electrolyte),","1 1"),((c),MgCl_(2)hArr Mg^(2+)+2Cl^(-)",",3" "(1-2x)),((d),Al(NO_(3))_(3)hArr Al^(3+)+3NO_(3)^(-)",",4" "(1-+3x)):}`

Discussion

2029.	For `0.1M` solution ,the colligative property will follow the orderA. `NaCl gt Na_(2)SO_(4)gt Na_(3)PO_(4)`B. `NaCl gt Na_(2)SO_(4)~~Na_(3)PO_(4)`C. `NaCl lt Na_(2)SO_(4) lt Na_(3)PO_(4)`D. `NaCl lt Na_(2)SO_(4)=Na_(3)PO_(4)`
Answer» Correct Answer - C Colligative property in decreasing order `Na_(3)PO_(4)gtNa_(2)SO_(4)gtNaCl` `Na_(3)PO_(4)rarr3Na^(+)+PO_(4)^(3-)=4` `Na_(2)SO_(4)rarr2Na^(+)+SO_(4)^(2-)=3` `NaClrarrNa^(+)+Cl^(-)=2`

Discussion

2030.	pH of a `0.1M`monobasic acid is found to be `2`,Hence its osmotic pressure at a given temp.TK is-A. `0.1RT`B. `0.11RT`C. `1.1RT`D. `0.01RT`
Answer» Correct Answer - B `underset(0.1)(HA)rarrH^(+)A^(-),pH=2` `pH=2`so `[H^(+)]=0.01` Total concentration =`0.1+0.01=0.11M` `pi=CRT=0.11RT`

Discussion

2031.	`pH` of a `0.1 M` monobasic acid is found to be `2`. Hence, its osmotic pressure at a given temperature T K isA. `1.1RT`B. `0.1RT`C. `0.11RT`D. `0.01RT`
Answer» Correct Answer - C `pH = 3` `pH = 2` `:.[H^(+)] = 10^(-2)` `C alpha = 10^(-2)` `alpha = (10^(-2))/(10^(-1)) = 10^(-1)` `i=1+alpha` `i=1 +0.1 = 1.1` `pi V = inRT` `pi =i((n)/(V))RT` `= 1.1 xx 0.1 xx RT = 0.11 RT`

Discussion

Explore topic-wise InterviewSolutions in .

The temperature at which molarity of pure water is equal to its molality is `:`A. 273 KB. 298 KC. 277 KD. None of these

Which is better method for expressing concentration of solution – Molarity or Molality

What are minimum boiling and maximum boiling azeotropes?

`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` barA. 0.94 barB. 9.4 barC. 0.094 barD. `9.4 xx 10^(-4)` bar

Define Azeotropes? What are maximum and minimum boiling azeotropes ? Explain with example.

`0.002 molar` solutiion of `NaCl` having degree of dissociation of `90%` at `27^(@)C` has osmotic pressure equal to a.0.94 bar , b.9.4 bar , c.0.094 bar , d.`9.4xx10^(-4)` bar

Calculate the amount of KCl which must be added to 1Kg of water so that the freezing point is depressed by 2 kelvin. (Kf for water = 1.86 K Kg mol-1)

Molarity and molality of a solution of caustic soda are respectively ` 11.12 M ` and ` 94.12 m ` The density of the solution isA. ` 0.556 g mL^(-1) `B. ` 5.56 g mL^(-1) `C. ` 55.6 gmL^(-1) `D. none of these

Molarity and molality of a solution of caustic soda are respectively ` 11.12 M ` and ` 94.12 m ` The density of the solution isA. `0.556 g mL^(-1)`B. `5.56 g mL^(-1)`C. `55.6 g mL^(-1)`D. None of these

Calcualate the amount of `NaCl` which must be added to `100 g` water so that the freezing point, depressed by `2 K`. For water `K_(f)` =`1.86 K kg mol^(-1)`.

10.0 grams of caustic soda when dissolved in 250 cm3 of water, the resultant molarity of solution is :(a) 0.25 M (b) 0.5 M(c) 1.0 M (d) 0.1 M

4.0g of caustic soda is dissolved in 100 cc of solution. The normality of solution isA. 1B. 0.1C. 0.5D. 4

Isotonic solutions have sameA. Molar concentrationB. MolalityC. NormalityD. None of these

A solution containing `0.85 g` of `ZnCI_(2)` in `125.0g` of water freezes at `-0.23^(@)C`. The apparent degree of dissociation of the salt is: `(k_(f)` for water `= 1.86 K kg mol^(-1)`, atomic mass, `Zn = 65.3` and `CI = 35.5)`A. `0.875 M`B. `1.00M`C. `1.75M`D. `0.975M`

A `0.01M` solution of glucose in water freezes at `-0.0186^(@)C`. A `0.01M` solution of KCI in water will freeze at temperatureA. higher than `-0.0186^(@)C`B. `0^(@)C`C. `0.0186^(@)C`D. lower than `-0.0186^(@)C`

Normal boiling point of water is `373 K`. Vapour pressure of water at `298 K` is `23 mm` enthalpy of vaporisation is `40.656 kJ mol^(-1)` if atmopheric pressure becomes `23 mm`, the water will boil at:A. `250K`B. `298K`C. `51.6k`D. `12.5K`

Normal boiling point of water is `373 K`. Vapour pressure of water at `298 K` is `23 mm` enthalpy of vaporisation is `40.656 kJ mol^(-1)` if atmopheric pressure becomes `23 mm`, the water will boil at:A. `250K`B. `294K`C. `51.6`D. `12.5K`

A `0.01m` aqueous solution of `K_(3)[Fe(CN)_(6)]` freezes ar `-0.062^(@)C`. What is the apparent percentage of dissociation? `[K_(f)` for water `= 1.86]`

0.01 molal aqueous solution of `K_(3)[Fe(CN)_(6)]` freezes at `-0.062^(@)C`. Calculate percentage dissociation `(k_(f)=1.86)`

A `0.2 molal` solution of `KCl` freezes at `-0.68^(@)C`. If `K_(f)` for `H_(2)O` is `1.86`, the degree of dissociation of `KCl` is a.`85%` , b.`83%` , c. `65%` , d. `90%`

A `0.2 molal` solution of `KCl` freezes at `-0.68^(@)C`. If `K_(f)` for `H_(2)O` is `1.86`, the degree of dissociation of `KCl` is a.`85%` , b.`83%` , c. `65%` , d. `90%`A. 0.75B. 0.83C. 0.65D. 0.92

`6xx10^(-3)` g oxygen is dissolved in per kg of sea water. Calculate the ppm of exgen in sea water.

A solution is prepared by dissolving 15g of cane sugar in 60g water. Compute the mass per cent of each component of solution.

A solution is prepared by dissolving 5.64g of flucose in 60g of water. Calculate the following : (i) mass percent of each of glucose and water (ii) molality of the solution (iii) mole fraction of each of glucose and water.

What is osmolarity of`0.20MKCl` solution?A. 0.10 osmolB. 0.20 osmolC. 0.30 osmolD. 0.40 osmol

Which has maximum osmotic pressure at temperature`TK`?A. 150ml of 1.0M urea solutionB. 210ml of 1.0M urea solutionC. Mixture of 150ml of 1.0M urea solution and 210 mL of 1.0M glucose solutionD. Allof the above are isotonic solutions

Which has the highest boiling point?A. `0.1m N_(2)SO_(4)`B. `0.1m Al(NO_(3))_(3)`C. `0.1m MgCl_(2)`D. `0.1m C_(6)H_(12)O_(6)("glucose")`

For `0.1M` solution ,the colligative property will follow the orderA. `NaCl gt Na_(2)SO_(4)gt Na_(3)PO_(4)`B. `NaCl gt Na_(2)SO_(4)~~Na_(3)PO_(4)`C. `NaCl lt Na_(2)SO_(4) lt Na_(3)PO_(4)`D. `NaCl lt Na_(2)SO_(4)=Na_(3)PO_(4)`

pH of a `0.1M`monobasic acid is found to be `2`,Hence its osmotic pressure at a given temp.TK is-A. `0.1RT`B. `0.11RT`C. `1.1RT`D. `0.01RT`

`pH` of a `0.1 M` monobasic acid is found to be `2`. Hence, its osmotic pressure at a given temperature T K isA. `1.1RT`B. `0.1RT`C. `0.11RT`D. `0.01RT`