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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Define Solutions. |
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Answer» Solutions are homogeneous mixtures of two or more substances (components). |
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| 2. |
At `27^(@)C`,3.92 gm `H_(2)SO_(4)` is present in 250 ml solution. The osmotic pressure of this solution is 1.5 atm. If the osmotic pressure of solution of NaOH is 2 atm at same temperature, then concentration of NaOH solution isA. 0.32 MB. 12.183 MC. `72.3 gm//lit`D. 1 M `Na_(2)SO_(4)` |
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Answer» Correct Answer - 1 |
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| 3. |
If K_(3)[Fe(CN)_(6)]` is completely ionised in a solution, what is the number of paritcles forms when one molecule is completely ionised. |
| Answer» `K_(3)[Fe(CN)_(6)]overset(("aq"))to3K^(+)(aq)+[Fe(CN)_(6)]^(3-)(aq).` | |
| 4. |
29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock solution required to prepare 200 mL of 0.4 M HCI |
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Answer» Mass of HCI=29.2g, Mass of solution = 100g `"Volume of solution"=((100g))/((1.25" g mL"^(-1)))=80mL=0.08 L` `M=("Mass of HCI/Molar mass")/("Volume of solution in litres")` `=((29.2g)//(36.5" g mol"^(-1)))/((0.08 L))=10M` `M_(1)V_(1)=M_(2)V_(2),(10M)xxV_(1)=(0.4M)xx(200 mL)` `V_(1)=((200mL)(0.4))/((10M))=8mL.` |
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| 5. |
A compound `H_(2)X` with molar mass of `80 g` is dissolved in a solvent having density of `0.4 g mL^(-1)`. Assuming no change in volume upon dissolution, the molality of a `3.2` molar solution isA. `9`B. `6`C. `5`D. `8` |
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Answer» Correct Answer - 4 Since these is no change in volume upon dissolution, the volume of solution is nearly equal to the volume of solvent. 3.2 molar solution implies that 3.2 moles of solute are present in every one litre of solution. Thus considering 1L of solution we have 3.2 moles of solute. Mass of solvent=(Volume of solvent)(Density of solvent) `=(1000 mL)(0.4 g mL^(-1))` `=400 g` `=0.4 kg` Thus, molarity (m)`=("moles of solute")/("mass of solvent" (kg))=(3.2 mol)/(0.4 kg)` `=8 mol kg^(-1)` |
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| 6. |
A compound `MX_(2)` with molar mass of 80 g is dissolved in a solvent having density 0.4 g `mL^(-1)`. Assuming that there is no change in volume upon dissociation, what is the molality of a 3.2 molar solution ? |
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Answer» 3.2 mola solution contains solute=3.2 mol Consider that the volume of solution =1L Since there is no change in volueme, on dissolution of the solvent, volume solvent = 1L Mass of the solvent = `vxxd` =`(100mL)xx(0.4 mg L^(-1))=400 g=0.4 kg` `("Molality of solution")=("No. of moles of solute")/("Mass of solvent in kg")` `=((3.2mol))/((0.4 kg))=8 "kg mol"^(-1)=8` molal |
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| 7. |
The solubility of a solute in a solvent (that is, the extent of the mixing of the solute and solvent species) depends onA. the natural tendency for the solute and solvent species to mixB. the natural tendency for a system to have the lowest energy possibleC. a balance between (1) and (2)D. neither (1) and (2) |
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Answer» Correct Answer - 3 The solubility of one substance in another can be explained in terms of two factors: One is the natural tendency to mix. This is sometimes referred to as the natural tendency toward disorder. Entropy is a measure of disorder. Thus, the tendency toward disorder can be expressed as a tendency toward increasing entropy. The mixing of substance increases entropy. This increse is one of the primary factors leading to the spontaneous formation of a solution. Consider a vessel which is divided by a removable partition with `O_(2)(g)` on the left and `N_(2)(g)` on the right. When the partition is removed, molecules of the two gases begin to mix. Ultimately, the molecules become throughly mixed through their random motion. We might expect a similar mixing of molecules or ions in other types of solutions. If the process of dissolving one substance in another involved nothing more than simple mixing, we would expect substances to be completely solube in one another, that is, we would expect substances to be miscible. We know that this is only sometimes the case. Usually, substance have limited solubility in one another. A factor that can limit solubility is the relative forces of attraction between species (molecules or ions). Suppose there are strong attraction solute species and strong attractions between solvent species, but weak attractions between solute and solvent species. In that case, the strongest attraction are maintained so long as the solute and solvent species do not mix. The lowest energy of the solute-solvent system is obtained then also. Thus, the solubility of a solute in a solvent depends on a balance between (1) and (2). As a general rule, "like dissolves like". Oil is miscible in gasoline a sboth are mixtures of hydrocarbons (compoinds of C and H only) while oil does not mix with water because water is a polar substances, whereas hydrocarbons are not. |
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| 8. |
Maximum amount of a solid aolute that can be dissolved in a specified amount of a given liquid solvent does not depend upon………..A. temperatureB. nature of soluteC. pressureD. nature of solvent |
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Answer» Correct Answer - C Maximum amount of solid that can be dissolved in a specified amount of a given solvent does not depend upon pressure. This is because sohlid and liquid are highly incompressible and practicully remain unaffected by change in pressure. |
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| 9. |
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent. It depends upon (i)nature of solute , (ii)nature of solvent , (iii)temperature , (iv)pressureA. only (i),(ii) and (iii)B. only (i), (iii) and (iv)C. only (i) and (iv)D. (i),(ii),(iii) and (iv) |
| Answer» Correct Answer - D | |
| 10. |
A complex is represented as `CoCl_(3) . XNH_(3)`. Its `0.1` molal solution in aqueous solution shows `Delta T_(f) = 0.558^(circ). (K_(f)` for `H_(2)O` is `1.86 K "molality"^(-1))` Assuming `100%` ionisation of complex and co-ordination number of `Co` as six, calculate formula of complex. |
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Answer» Correct Answer - `[Co(NH_(3))_(5)Cl]Cl_(2)` `Delta"T"_("f")=[1+("n"-1)alpha]"K"_("f")xx"Molality"` `0.558=[1+("n"-1)xx1]xx1.86xx0.1` `"n"=(0.558)/(1.86xx0.1)` `"n"=3` `therefore "formula of complex"=["CO"("NH"_(3))_(5)"Cl"]"Cl"_(2)` |
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| 11. |
Assertion:The boiling and melting point of amides are higher than corresponding acids. Reason:It si due to strong intermolecular hydrogen bonding in their molecules.A. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both the assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - A The boiling point and melting point are higher due to presence of the intermolecular hydrogen bonding. |
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| 12. |
A complex is represented as `CoCl_(3)Xnh_(3)` . Its `0.1` molal solution in water `DeltaT_(f)=0.588 K. K_(f)` for `H_(2)O` is `1.86K" molality"^(-1)`. Assuming `100%` ionisation of complex and co`-` ordination number of `Co` is six calculate formula of complex. |
| Answer» `[Co(NH_(3))_(5)Cl]Cl_(2)` | |
| 13. |
`20`g of a binary electrolyte(mol.wt.=`100`)are dissolved in `500`g of water.The freezing point of the solution is `-0.74^(@)CK_(f)=1.86K "molality"^(-1)`.the degree of ionization of the electrolyte isA. `50%`B. `75%`C. `10%`D. `0%` |
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Answer» Correct Answer - D `DeltaT=(1000xxK_(f)xxw)/(mxxw)` `:.0.74=(1000xx1.86xx20)/(mxx500)` `:.m=100` Now,`("normal,mol .wt.")/("exp.mol.wt.")=1+alpha` `:.(100)/(100)=1+alpha:.alpha=0` |
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| 14. |
36g of glucose (molar mass = `180 g//mol)` is present in 500g of water, the molarity of the solution isA. 0.2B. 0.4C. 0.8D. 1 |
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Answer» Correct Answer - B Molarity `= (W xx 1000)/(M xx W)` where V = volume of water `= 500 mL` W = mass of glucose `= 36g` M = mol mass of glucose `= 180 g mol^(-1)` Molarity `= (36 xx 1000)/(180 xx 500) = 0.4M` |
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| 15. |
What will be the molality of a solution having 18g of glucose (Mol. wt 180) dissolved in 500g of water?A. 1mB. 0.5mC. 0.2mD. 2m |
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Answer» Correct Answer - C Molality `= (18)/(180) xx (1000)/(500) = 0.2m` |
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| 16. |
What is “semi permeable membrane”? |
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Answer» Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through which small solvent molecules like water can pass; but the passage of bigger molecules of solute is hindered, are known as semi permeable membrane. |
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| 17. |
How does sprinkling of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process. |
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Answer» When salt is spread over snow covered roads, snow starts melting from the surface because of the depression in freezing point of water and it helps in clearing the roads. |
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| 18. |
How does sprinking of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process.A. lowering in vapour pressure of snowB. depression in freezing point of snowC. increase in freezing point of snowD. melting of ice due to increase in temperature by putting salt |
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Answer» Correct Answer - B When salt is spread over snow, snow starts melting from the surface because of depression in freezing point and helps in clearing the roads . |
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| 19. |
Assertion (A): Sodium chloride used to clear snow on the roads. Reason (R ): Sodium chloride depresses the freezing point of water.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explanation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is correct, but `(R )` is correct. |
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Answer» Correct Answer - A According to colligative property addition of non-volatile solute to volatile solvent, the freezing point of solution decreases. |
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| 20. |
How does sprinking of salt help in clearing the snow covered roads in hilly areas? Explain the phenomenon involved in the process. |
| Answer» The phenomenon involed in the melting of snow in snow coverdroads in the depression in freezing point which is caused by7 the addintion of non-volatile impureities to a liquid. Addition of salt (sodium chloride) lowers the frzzing point temperature of water and thus, helps in the melting of snow, For details, conslt section 11. | |
| 21. |
Sodium choride or calcium chloride is used to clear snow from the roads. Why? |
| Answer» Sodium chloride depress the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature. | |
| 22. |
Osmotic pressure of an aqueous solution at `27^(@)C` is found to be 1900 mm Hg. What will be the freezing point of solution (Assuming, molality =molarity xx1.5) ? `(K_(f)-=1.86)` |
| Answer» Correct Answer - `-0.283^(@)C` | |
| 23. |
For a dilute solution containing a nonvolatile solute, the molar mass of solute evaluated from the elevation of boiling point is given by the experssion:A. `M_(2) = (DeltaT_(b))/(K_(b)) (m_(1))/(m_(2))`B. `M_(2)=(DeltaT_(b))/(K_(b))(m_(2))/(m_(1))`C. `M_(2) =(K_(b))/(DeltaT_(b)) (m_(2))/(m_(1))`D. `M_(2) = (K_(b))/(DeltaT_(b)) (m_(1))/(m_(2))` |
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Answer» Correct Answer - C `DeltaT_(b) = (K_(b)m_(2))/(M_(2)m_(1))` |
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| 24. |
A solution of `0.450g` of urea (mol.wt 60) in `22.5g` of water showed `0.170^(@)C` of elevation in boiling point, the molal elevation constant of water:A. `0.51`B. `0.95`C. `0.25`D. `2.25` |
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Answer» Correct Answer - A `0.17 = ((0.45)/(60))/(22.9) xx 1000 xx K_(b)` `(0.17)/(0.33) = K_(b), K_(b) = 0.51` |
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| 25. |
`0.01(M)` solution an acid HA freezes at `-0.0205^(@)C`. If `K_(f)` for water is `1.86K kg mol^(-1)`, the ionization constant of the conjugate base of the acid will be (consider molarity)`~=` molarity)A. `1.1xx10^(-4)`B. `1.1xx10^(-3)`C. `9xx10^(-11)`D. `9xx10^(-12)` |
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Answer» Correct Answer - C `DeltaT_(f)=k_(f)xxm=1.86xx0.01=0.0186` `i=(0.0205)/(0.0186)=1.10=1+alpha`or`alpha=0.1` `K_(a)=((Calpha)^(2))/(1-alpha)=(0.01xx(0.1)^(2))/(1-0.1)=(1)/(9)xx10^(-3)` or `K_(b)=9xx10^(-11)` |
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| 26. |
An aqueous solution of `NaCI` freezes at `-0.186^(@)C`. Given that `K_(b(H_(2)O)) = 0.512K kg mol^(-1)` and `K_(f(H_(2)O)) = 1.86K kg mol^(-1)`, the elevation in boiling point of this solution is:A. `0.0585 K`B. `0.0512 K`C. `1.864 K`D. `0.0265K` |
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Answer» Correct Answer - B `0.186 = 1.86 xm` `m = 0.1` `DeltaT_(b) = 0.1 xx 0.512 = 0.0512` |
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| 27. |
Observe the P-T phase diagrams for a given substance A. Then melting point of A(s) ,boiling point of `A(l)`,critical point of A and triple point of A (at their respective pressure) are respectively- A. `T_(1),T_(2),T_(3),T_(4)`B. `T_(4),T_(3),T_(1),T_(2)`C. `T_(3),T_(4),T_(2),T_(1)`D. `T_(2),T_(1),T_(3),T_(4)` |
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Answer» Correct Answer - C `mpT_(3) rarr`solid &liquid ,bp`T_(4)rarr`liquid - gas Critical point `T_(2)`, triple point `T_(1)rarr`solid liquid gas |
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| 28. |
An aqueous solution of `2` per cent `(wt.//wt)` non-volatile solute exerts a pressure of `1.004` bar at the boiling point of the solvent. What is the molecular mass of the solute?A. `"23.4 g mol"^(-1)`B. `"41.35 g mol"^(-1)`C. `"10 g mol"^(-1)`D. `"20.8 g mol"^(-1)` |
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Answer» Correct Answer - B Vapour pressure of pure water at boiling point =1 atm = 1.013 bar Vapour pressure of solution `(p_s)`=1.004 bar Let mass of solution = 100 g Mass of solute=2 g Mass of solvent = 100-2= 98 g `(p^@-p_s)/p^@=n_2/(n_1+n_2)=n_2/n_1 =(W_2//M_2)/(W_1//M_1) ( because n_2 lt lt lt n_1)` `(1.013-1.004)/1.013=2/M_2xx18/98` or `M_2=(2xx18)/98xx(1.013)/0.009 = 41.35 g mol^(-1)` |
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| 29. |
How many grams of KCl should be added to 1kg of water to lower its freezing point to ‐8.0 °C (kf = 1.86 K kg /mol) |
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Answer» Since KCl dissociate in water completely L = 2 ΔTf = ikf x m; m = ΔTf/ikf m = 8 / 2 x 1.86 = 2.15 mol/kg. Grams of KCl = 2.15 x 74.5 = 160.2 g/kg. |
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| 30. |
Find the freezing point of the solution containing 3.6 g of glucose dissolved in 50 g of H2O. (Kf for H2O = 1.86 K/m). |
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Answer» Mass of glucose, WB = 3.6 g Molecular mass of glucose = 180 g mol-1 Mass of solvent, WA = 50 g Kf for H2O = 1.86 K/m = 1.86 K kg-1 mol-1 MB = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,\bigtriangleup T_f}\) ∴ \(\bigtriangleup T_f\) = \(\frac{1000K_f\,\times\,W_B}{W_A\,\times\,M_B}\) = \(\frac{1000\,g\,kg^{-1}\,\times\,1.86\,K\,kg^{-1}\,mol^{-1}\,\times\,3.6\,g}{50\,g\times180\,g\,mol^{-1}}\) = 0.744 K. i.e., ΔTf = T°f – Tf = 0.744 K ∴ Freezing point of the solution, Tf = T°f – ΔT = 273 K – 0.744 K = 272.3 K |
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| 31. |
A solution of sucrose (molar mass =342) is prepared by dissolving 688.4 g in 1000 g of water. Calculate The freezing point of solution.A. `273`B. `373`C. `272.628`D. `271.628` |
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Answer» Correct Answer - C Freezing point of solution can be calculated by using the relation `DeltaT_(f)=(K_(f) xx 1000 xx W_(B))/(W_(A) xx Mw_(B))` `K_(f)=1.86 K m^(-1)` `:. Delta T_(f)=(1.86 xx 1000 xx68.4)/(1000 xx 342)=0.372` Freezing point fo solution =Freezing point of pure water -`DeltaT_(f)` =273-0.372=272.628 K |
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| 32. |
Solution of sucrose is prepared by dissolving 34.2 g of it in 1000 g of water. Find out the freezing point of the solution, if Kf of water is 1.86 K/kg/mol? (Molecular mass of sucrose is 342 g/mol). |
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Answer» ΔTf = kf × m = 1.86 K kg mol-1 × \(\frac{34.2\times1000g\,kg^{-1}}{342g\,mol^{-1}\times\,1000g}\) = 0.186 K Tf = OK – 0.186 K Freezing point of the solution = – 0.186 K |
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| 33. |
Elevation in boiling point of an aqueous solution of a non-electroluyte solute is `2.01^(@)` . What is the depression in freezing point of the solution ? `K_(b)(H_(2)O)=0.52^(@)mol^(-1)kg` `K_(f)(H_(2)O)=1.86^(@)mol^(-1)kg`A. `7.17^(@)`B. `0.52^(@)`C. `3.57^(@)`D. `0.93^(@)` |
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Answer» Correct Answer - A `DeltaT_(b)` = Molarity `xx K_(b)` `DeltaT=mK_(b)` `DeltaT_(f)=mK_(f)` `:. (cancelOT_(f))/(cancelOT_(b))=(K_(f))/(K_(b))=(1.86)/(0.52)=3.57` |
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| 34. |
The density of a solution prepared by dissolving 120 g of urea (mol. Mass=60 u) in 1000 g of water is 1.15 g/mL. The molarity if this solution isA. 1.78 MB. 1.02 MC. 2.05 MD. 0.50 M |
| Answer» Correct Answer - C | |
| 35. |
The density of a solution prepared by dissolving 120 g of uree (molar mass = 60 u) in 1000 g of water is 1.15 g/mL The molarity of solution is:A. 0.50 MB. 1.78 MC. 1.02 MD. 2.05 M |
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Answer» Correct Answer - d Mass of solution=(1000+120)=1120 g Volume of solution =`((1120g))/(115 g mL^(-1))=973.9 mL` `"Molarity (M)=((120g)//(60 g mol))/(973.9//(1000K))` `=2.05 molL^(-1)=2.05 M`. |
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| 36. |
A solution is prepared by dissolving 10 g NaOH in 1250 ml of a solvent of density 0.8 ml/g. The molarity of the solution is _______ |
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Answer» A solution is prepared by dissolving 10 g NaOH in 1250 ml of a solvent of density 0.8 ml/g. The molarity of the solution is 0.25 |
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| 37. |
n moles of a non volatile solute are dissolved in wg of water. If `K_(f)` is the molal depression constant of water, the freezing point of the solution will beA. `(1000 K_(f)n)/(w)`B. `-(1000K_(f)n)/(w)`C. `(1000K_(f)w)/(n)`D. `-(1000K_(f)w)/(n)` |
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Answer» Correct Answer - A We know that `DeltaT_(f) = (k_(f)xxw xx 1000)/(MxxW)` ltbr. Or `T^(-)-T =(k_(f)xxwxx1000)/(MxxW)` |
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| 38. |
A solution of HCI is prepared by dissolving 5.5 g HCI in 200 g ehanol. The density of the solution is 0.79 g `mL^(-1)` Molarity of solution is :A. 0.58 MB. 0.16 MC. 0.92 MD. 1.2 M |
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Answer» Correct Answer - a `"Molarity (M)"=("Mass of HCI//Molar mass")/("Volame of solution in litrese")` Mass of HCI solution = 200+5.5=205.5 g Density of solution = 0.79 g `mL^(-1)` `"Volume of solution"=((205.5g))/((0.79gmL^(-1)))` =260L `"Molarity"=((5.5g)//(36.5g mol^(-1)))/(0.260L)` =0.58 `mL^(-1)=0.58 M` |
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| 39. |
Y g of non-volatile organic substance of molecular mass M is dissolved in 250 g benzene . Molal elevation constant of benzene is `K_(b)` . Elevation in its boiling point is given by :A. `(M)/(K_(b)y)`B. `(4K_(b)y)/(M)`C. `(K_(b)y)/(4M)`D. `(K_(b)y)/(M)` |
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Answer» Correct Answer - B `Delta T = K_(b) xx (w_(B) xx 1000)/(m_(B) xx w_(A))` `= K_(b) xx (y xx 1000)/(M xx 250) = (4 K_(b) y)/(M)` |
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| 40. |
Figure explains elevation in boiling point when a non-volatile solute is added to a solvent. Provide a thermodynamic explanation of the elevation in boiling point. |
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Answer» Correct Answer - At the boiling point liquid and vapour pase are in equilibrium. `therefore DeltaG=0` Also `DeltaG=DeltaH-TDeltaS` `therefore T=(DeltaH)/(DeltaS)` If we compare the heats of vaporisation `(DeltaH)~ for a pure solvent and for a solution, we find that the two values are simi9lar because similar intermolecular forces between solvent molecule must be overcome in both cases. In comparing the entropies of vaporisation however, the two values are not similar. because a solution has more molecular randomness than a pure solvent has, the entropy change between solution and vapour is smaller than the entropy change betweeen pure solvent and vapour. Thus, `T` (boiling of solution) must be higher than that of pure solvent. |
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| 41. |
Y g of a non-volatile organic subsatance of molecular mass M is dissoved in 250 g of benzene. Molal elevation constant of benzene is `K_(b)`. Elevation in boiling point is given byA. `M/(K_(b)Y)`B. `(4K_(b)Y)/M`C. `(K_(b)Y)/(4M)`D. `(K_(b)Y)/M` |
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Answer» Correct Answer - b `DeltaT_(b)=K_(b)xx(W_(B)xx1000)/(M_(B)xxW_(A))` `=K_(b)xx((Y_(g))xx(1000 g ))/(Mxx(250g ))` `(K_(b)xxYxx4)/M=(4K_(b)y)/M` |
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| 42. |
Calculate the mass of NaCI (molar mass =58.5 g `mol^(-1)`) to be dissoved in 37.2 g of water to lower the freezing point by `2^(@)C` assuming that NaCI underfoes complete dissocilation in aqueous solution (`K_(f)` for water = 1.86 K kg `mol^(-1)`). |
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Answer» Correct Answer - 1.16 g `i=(DeltaT_(f))/(K_(f)xxm)=(DeltaT_(f)xxM_(B)xxM_(A))/(K_(f)xxW_(B))or W_(B)=(DeltaT_(f)xxM_(B)xxM_(A))/(ixxK_(f))` `i=2("for complete dissciation of NaCI"),K_(f)=1.86" K kg mol"^(-1)` `DeltaT_(f)=2K, M_(B)=58.5" g mol"^(-1), M_(A)=37.2 g=0.037 kg` `W_(B)=((2K)xx(58.5"gmol"^(-1))xx(0.037 kg))/(2xx(1.86" K kg mol"^(-1)))=1.16 g` |
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| 43. |
When `0.004 M Na_(2)SO_(4)` is an isotonic acid with `0.01 M `glucose, the degree of dissociation of `Na_(2)SO_(4)` isA. `75%`B. `85%`C. `50%`D. `25%` |
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Answer» Correct Answer - 1 Isotonic means their effective molarition are equal: `(i C)_(Na_(2)SO_(4))=(I C)_("glucose")` `i_(Na_(2)SO_(4))=((I C)_("glucose"))/C_(Na_(2)SO_(4)` `=((1)(0.01 M))/((0.004 M))` `=2.5` `{:(,Na_(2)SO_(4)(aq)hArr,2Na^(+)(aq),+,SO_(4)^(2-)(aq)),("Moles before dissociation",1mol,0mol,,0mol),("Moles before dissociation",(1-alpha)mol,2alphamol,,alphamol):}` `i=("Total moles of particles after dissociation")/("Total moles of particles before dissociation")` `i=((1-alpha)+(2alpha)+(alpha))/1` `i=1+2alpha` `alpha=(i-1)/2=(2.5-1)/2` `=0.75` Thus, percent dissociation`=0.75xx1.007`. |
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| 44. |
Equimolal solutions `A` and `B` show depression in freezing point in the ratio `2:1`. A remains in the normal state in solution. `B` will beA. Normal in solutionB. Dissociated in solutionC. Associated in solutionD. Hydrolysed in solution |
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Answer» Correct Answer - C `(DeltaT_(f _A)) /(DeltaT_(f_B))=2/1 = 1/(1//2)` , i.e., `B` should associate to show higher `DeltaT`. |
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| 45. |
Equimolal solutions `KCl` and compound `X` in water show depression in freezing point in the ratio of `4:1`, Assuming `KCl` to be completely ionized, the compound `X` in solution mustA. Dissociate to the extent of `50%`B. Hydrolyze to the extent of `80%`C. Dimerize to the extent of `50%`D. Trimerize to the extent of `75%` |
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Answer» Correct Answer - D `underset(("p moles"))(KCl) and underset(("p moles"))(X)` `DeltaT_(f) (KCl)=iK_(f)m=2K_f m` `DeltaT_f (X) = iK_f .m=1/2(2K_f m)rArr i=1/2 (lt1)` `2X hArr X_2` `{:(1,0),(1-alpha , a//2):}` `i=1+alpha+alpha/3=1-(2alpha)/3=1/2rArr alpha=3/4=75%` |
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| 46. |
Which of the following `0.1 M` aqueous solutions will have the lowest freezing point?A. Potassium sulphateB. Sodium chlorideC. UreaD. Glucose |
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Answer» Correct Answer - A Urea and glucose do not dissociate in solution. Sodium chloride gives two ions and potassium sulphate gives three ions per formula unit. Therefore, the effective number of particles is maximum in potassium sulphate, and it shows the maximum depression in freezing point. |
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| 47. |
The freezing point among the following equimolal aqueous solutions will be highest forA. `C_(6)H_(5)NH_(3)Cl`(aniline hydrochloride)B. `Ca(NO_(3))_(2)`C. `La(NO_(3))_(3)`D. `C_(6)H_(12)O_(6)`(glucose) |
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Answer» Correct Answer - D `C_(6)H_(5)NH_(3)Cl rarr C_(6)H_(5)NH_(3)^(o+)+Cl^(ө)` `Ca(NO_(3))_(2) rarr Ca^(2+) + 2NO_(3)^(ө)` `La(NO_(3))_(3) rarr La^(3+) + 3NO_(3)^(ө)` Glucose does not dissociate. So it has the minimum number of particles, and therefore, it shows minimum depression in freezing point. So it has the maximum freezing point. |
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| 48. |
In a pair of immiscible liquid, a common solute dissolves in both and the equilibrium is reached. Then, the concentration of the solute in upper layer isA. in fixed ratio with that in the lower layerB. same as the lower layerC. lower than the lower layerD. higher than the lower layer |
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Answer» Correct Answer - A According to Nernst distribution law when we mixed a common solute in a pair of immiscible liquids, then the ratio of amount of solute in both liquids is fixed temperature. |
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| 49. |
In a pair of immiscible liquida, a common solute dissolves in both and the equilibrium is reached. Then, the concentration of the solute in upper layer isA. lower than the lower isB. in fixed ratio with that in the lower layerC. higher than the lower layerD. same as the lower layer |
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Answer» Correct Answer - 2 According to the Nernst distribution or partition law, if to a mixture of two immiscible liquids a substance (solute) which is soluble in both is added in varying amounts, then experiment shows that it distributes itself in such a way that the ratio of its concentration in the two layers is constant at constant temperature, i.e. `("Conc. of solute in liquid A")/("Conc. of solute in liquid B")=" constant", k` where the constant k is called the distribution or partition coefficient of that system at that temperature. The distribution law only holds true if the solute is in the same molecular condition in both the solvent. Thus, the ratio of the concentration yakes place to some extend in one solvent only and not in the other. |
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| 50. |
The freezing point of a solution prepared from `1.25 g` of non-electrolyte and `20 g` of water is `271.9 K`. If the molar depression constant is `1.86 K mol^(-1)`, then molar mass of the solute will beA. 105.7B. 106.7C. 115.3D. 93.9 |
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Answer» Correct Answer - A `M = (K_(f) xx 1000 xx w)/(W xx DeltaT_(f))` `= (1.86 xx 1000 xx 1.25)/(20 xx 1.1) = 105.68 ~~ 105.7` |
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