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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.
| 901. |
Define Azeotropes? |
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Answer» Constant boiling mixtures are called Azeotropes. |
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| 902. |
What do you understand by saying that molality of a solution is 0.2? |
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Answer» This means that 0.2 mol of the solute is dissolved in 1Kg of the solvent. |
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| 903. |
A solution contains 0.8960 g of K2SO4 in 500 ml solution. Its osmotic pressure is found to be 0.690 atm at 27°C. calculate the value of vant hoff factor. |
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Answer» MB = WB RT/ ∏ V WB = 0.8960 g V = 500 ml pA0 = 0.5 L R = 0.082 L atm K-1 mol-1 ∏ = 0.69 atm T = 300 K M = 0.896 g x 0.082 L atm K-1 mol-1 x 300 K = 0.69 atm x 0.5 L = 63.9 g mol-1 Normal molar mass = 2 x 39 + 32 + 4 x 32 = 174 g mol-1 molar mass/ Observed molar mass = 174 g mol-1 = 2.72 = 63.9 g mol-1 |
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| 904. |
Calculate the molarity of pure water (d=1/mL). |
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Answer» Consider 1000 mL of water Mass of 1000 mL of water `= 1000 xx 1 = 1000` g No. of moles of water `= (1000)/(18)=55.5` Molarity `= ("No. of moles of water")/("Volume in litres")` `=(55.5)/(1)=55.5M`. |
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| 905. |
Compared to common colloidal sols milcells have:A. higher colligative propertiesB. lower colligative propertiesC. same colligative propertiesD. None of these |
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Answer» Correct Answer - B Micelles have large molar mass so less colligative property. |
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| 906. |
Ideal solution Positive deviationNegative deviation |
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Answer» Examples of Ideal Solutions
Examples of Positive deviation
Examples of Negative deviation
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| 907. |
Molar massMolarity Number of molesMole fraction |
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Answer» Intensive Properties Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include: Boiling Point, Density, State of Matter, Color, Melting Point, Odor, Temperature, Refractive Index, Luster, Hardness, Ductility, Malleability Extensive Properties Extensive properties do depend on the amount of matter that is present. An extensive property is considered additive for subsystems. Examples of extensive properties include: Volume, Mass, Size, Weight, Length Extensive properties Mass, internal energy, heat capacity. Intensive properties Pressure, molar heat capacity, density, mole fraction, specific heat, temperature and molarity. Ratio of two extensive properties is always intensive. Extensive/Extensive = Intensive So, mole fraction and molarity are intensive properties. |
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| 908. |
What will be the molarity of 30 mL of 0.5 M `H_2SO_4` solution diluted to 500 mL ?A. 0.3 MB. 0.03 MC. 3 MD. 0.103 M |
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Answer» Correct Answer - B `V_1`=30 mL , `M_1`=0.5 M, `V_2`=500 mL, `M_2`=? , `M_1V_1=M_2V_2` 0.5 x 30 = `M_2 xx500` or `M_2`=0.03 M |
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| 909. |
Calculate the concentration of NaOH solution g/mL, which has the same normality as that of a solution of `HCl` of concentration 0.04 g/mL. |
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Answer» `N_(HCl)=(w_(B)xx1000)/(E_(B)xxV)=(0.04xx1000)/(36.5xx1)=1.095` `N_(NaOH)-=N_(HCl)` `:. 1.095=(w_(B)xx1000)/(40xx1)` `w_(B)=0.0438g//mL`. |
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| 910. |
How many `Na^(+)` ions are present in 50mL of a 0.5 M solution of `NaCl` ?A. 0.125 mol , 7.32 gB. 7.32 mol , 0.125 gC. 0.125 mol, 0.125 gD. 7.32 mol , 7.32 g |
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Answer» Correct Answer - A Number of moles `= (MV)/(1000)=(0.5xx250)/(1000)=0.125` Mass of NaCl `=58.5xx0.125=7.32g`] |
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| 911. |
How many `Na^(+)` ions are present in 50mL of a 0.5 M solution of `NaCl` ? |
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Answer» Number of moles of `NaCl = (MV)/(1000)` `= (0.5 xx 50)/(1000)=0.025` `NaCl rarr Na^(+) +Cl^(-)` Number of moles of `Na^(+)=` Number of moles of NaCl `= 0.025` Number of ions of `Na^(+)=0.025 xx 6.023 xx 10^(23)` `= 1.505 xx 10^(22)`. |
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| 912. |
How many `Na^+` ions are present in 100 mL of 0.25 M of NaCl solution ?A. `0.025 xx 10^23`B. `1.505xx10^22`C. `15xx10^22`D. `2.5xx10^23` |
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Answer» Correct Answer - B No. of moles of NaCl =`(MxxV)/1000=(0.25xx100)/(1000)=0.025` `NaCl to Na^(+) + Cl^(-)` No. of moles of `Na^+` ions = 0.025 No. of `Na^+` ions = 0.025 x 6.023 x `10^23` = `1.505xx10^22` |
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| 913. |
How many moles of sodium chloride present in 250 mL of a 0.50 M NaCl solution ?A. 7.32 gB. 3.8 gC. 5 gD. 0.5 g |
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Answer» Correct Answer - C No. of moles of NaOH =`(MxxV)/1000=(0.5xx250)/1000`=0.125 Mass of NaOH= 40 x 0.125 = 5 g |
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| 914. |
What type of liquids form ideal solutions? |
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Answer» Liquids having similar structures and polarities. |
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| 915. |
What is van’tHoff factor for KCl in aqueous solution? |
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Answer» i = 2 is van’tHoff factor for KCl in aqueous solution. |
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| 916. |
An alloy of copper and aluminium is prepared from 65.6 g of Cu and 423.1 g of Al. Calculate the mass percent of each component in this solid solution. Strategy: The mass precentage of a component of a solution is define as: Mass% of a component =`("Mass of the component in the solution")/("Total mass of the solution")xx100%` |
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Answer» The data tell us that the total mass of the solution is 65.6 g of `Cu+423.1 g` of `Al=488.7 g` of solution and that there is `65.6 g` of Cu in the solution. Therefore: Mass percent of `Cu=(65.6 g Cu)/(488.7 g soln)xx100%` `=13.4%` Mass percent of `Al=(100%)-(13.4%)` `=86.6%` |
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| 917. |
Why does chemist prefer to refer concentration of solution in terms of molality? (Or)Which of the two molality and molarity is better to express concentration of solution? Why? |
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Answer» Molality does not change with temperature where as molarity changes with temperature because volume changes with temperature. Therefore molality is better. |
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| 918. |
A chemist prepared a solution by adding by `230 g` of pure ethanol `(C_(2)H_(5)OH)` to `144 g` of water. Calculate the mole fractions of these two components. The molar masses of ethanol and water are `64 g` and `18 g`, respectively. Strategy: To calculate the mole fractions, we must find the number of moles of ethanol and of water. To find the moles, we must divide the given mass by the respective molar mass. In a two-component system made up of `A` and `B` molecules, the mole fraction of a component of a solution, say, component A is written `chi_(A)` and is defined as Mole fraction of component `A= chi_(A)` `=("moles of A")/("sum of moles of A and B")` |
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Answer» The number of moles of `C_(2)H_(5)OH and H_(2)O` present in solution are Moles of ethnol `(n_(C_(2)H_(5)OH))= ("mass of ethanol")/("molar mass of ethanol")` `=(230 g C_(2)H_(5)OH)/(46 g C_(2)H_(5)OH//"mole" C_(2)H_(5)OH)` `=5 mol C_(2)H_(5)OH` Moles of water `(n_(H_(2)O))=("mass of water")/("molar mass of water")` `=(144 g H_(2)O)/(18 g H_(2)O //"mol" H_(2)O)` `=8 "mol" H_(2)O` Using Equation (2.7), we can write the mole fractions of ethanol and water as `chi_(C_(2)H_(5)OH)=n_(C_(2)H_(5)OH)/(n_(C_(2)H_(5)OH)+n_(H_(2)O))` `=(5 mol)/((5+8) mol)` `chi_(H_(2)O)=n_(H_(2)O)/(n_(C_(2)H_(5)OH)+n_(H_(2)O))` `=(8 mol)/((5+8) mol)=0.6154` |
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| 919. |
What is the molarity of an aqueous solution of NaOH contaiining 0.5 g in 500 `cm^(3)` of the solution ? |
| Answer» `"Molarity (M)"=((0.5g)//(40g mol^(-1)))/((0.5 dm^(3)))=0.025 moldm^(-3)(0.025 M)`. | |
| 920. |
Calculte the number of molecules of molecules of oxalic acid in 100 mL of 0.02 N oxalic acid soliution. |
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Answer» Correct Answer - `6.022xx10^(20)` Step I. Calculation of no. of moles of oxalic acid per litre of the solution. `"Molarity"=("No. of moles of oxalic acid")/("Volume of solution in litre")` `(0.01" mol L"^(-1))=("No. of moles of oxalic acid")/(1L)` No. of moles = 0.01 mol. Step II. Calculation of no. of molecules of oxalic acid per 100 mL of solution. 1000 mL of oxalic acid solution contains moles = 0.01 `"100 mL of oxalic acid solution contains moles"=(0.01)/(1000)xx100=0.001 mol` `therefore "Molecules of oxalic acid"=0.001xxN_(0)=10^(-3)xx6.022xx10^(23)=6.022xx10^(20)" molecules".` |
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| 921. |
A solution of urea in water boills at `100^(@)C`. Calculate the freezing point of the same solution. Molal constant `K_(f) and K_(b)` are 1.86 and 0.521 k `m^(-1)` repectivly. |
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Answer» Correct Answer - `-0.654^(@)C` `DeltaT_(f)=K_(f)xxm" and "DeltaT_(b)xxm` `(DeltaT_(f))/(DeltaT_(b))=K_(f)/K_(b)` `DeltaT_(b)=100.18^(@)C-100.0=0.18^(@)C=0.18 K` `K_(f)=1.86" K m"^(-1), K_(b)=0.512" K m"^(-1)` `DeltaT_(f)=K_(f)/K_(b)xxDeltaT_(b)=((1.86"Km"^(-1)))/((0.512"Km"^(-1)))xx0.18K=0.654K=0.654^(@)C` `therefore "Freezing point of soluion"=0^(@)C-0.654^(@)C=-0.654^(@)C.` |
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| 922. |
A solution is made by dissoving 40.5 g of `C_(2)H_(5)OH` until the total volume of the solution is 100 mL. What is the percentage volume of `C_(2)H_(5)OH`? The density of pure `C_(2)H_(5)OH` is 0.78 g/mL. |
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Answer» Correct Answer - 51.92% m `"Intial mass of "C_(2)H_(5)OH=40.5 g.` `"Density of "C_(2)H_(5)OH=0.78 g//mL.` `"Intial volume of solution "="Mass"/"Density"=((40.5g))/((0.78g//mL))=51.92 mL` Final volueme of solution = 100 mL `"Precentage volume of "C_(2)H_(5)OH=("Intial volume")/("Final volume")xx100=((51.92mL))/((100mL))xx100=51.92%` |
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| 923. |
The molal depression constant for water is `1.86^(@)C`. The freezing point of a `0.05-molal` solution of a non-electrolyte in water isA. `-1.86^(@)C`B. `-0.93^(@)C`C. `0.093^(@)C`D. `0.93^(@)C` |
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Answer» Correct Answer - C `DeltaT_(f)=K_(f) xx "Molality" =1.86 xx 0.05 =0.093` Thus, freezing point =`1-0.093 = -0.093` |
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| 924. |
A cane sugar solution has an osmotic pressure of `2.46` atm at `300K`. What is the strength of the solution |
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Answer» `pi V = nST` or `pi = (n)/(V) ST = CST` or `C =(pi)/(ST) = (2.46)/(300 xx 0.0821) = 0.1M` |
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| 925. |
The molal depression constant for water is `1.86^(@)C`. The freezing point of a `0.05-molal` solution of a non-electrolyte in water is |
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Answer» `Deltat_(f) =` molability `xx K_(f)` `= 0.05 xx 1.86 = 0.093^(@)C` `(T_(f))_(s) = T_(0) -0.093 = 0-093` `(T_(f))_(s) = -0.093^(@)C` |
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| 926. |
The number of mole of `KMnO_(4)` that will be needed to react completely with one mole of ferrous oxalate in acidic solution is:A. 0.2 molesB. 0.6 molesC. 0.4 molesD. 7.5 moles |
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Answer» Correct Answer - 2 `underset("Ferrous oxalate")(Fe(COO)_(2)) rarr Fe^(2+)+(COO^(-))_(2)` `underset(O.N=+2)(Fe^(2+)) overset(MnO_(2)^(+))(rarr) underset(O.N=+3)(Fe^(3+))` In the first reaction, O.N changes by 1 unit while in the second reaction O.N changes by 2 units. Thus, total change in O.N is by 3 units. This implies the one mole of ferrous oxalate contains 3 equivalents. `underset(Mn=+7)underset(O.N of)(MnO_(4)^(-))rarr underset(Mn=+2)underset(O.N. of)(Mn^(2+)) ("acidic medium")` Since in the above reaction, O.N. of Mn changes by 5 units, one mole of `MnO_(4)^(-)` contains 5 equivalents. According to the law of equivalence, we need 3 equivalents of `MnO_(4)^(-)` to oxidize one mole of ferrous oxalate completely in acidic medium. Since 5 equivalents of `MnO_(4)^(-)` come from 1 mole, 3 equivalents of `MnO_(4)^(-)` will come from `=1/5xx3` `=0.6 mol` |
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| 927. |
Assertion:For isotonic solution`C_(1)=C_(2)` Reason:For isotonic solution`pi_(1)=pi_(2)`.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 - D For isotonic solution osmotic pressure are same. Concentrations are same only when solute neither dissociated nor associates. |
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| 928. |
Assertion: Ebullioscopy or cryoscopy cannot be used for the determination of molar mass of polymers. Reason:High molar mass solute leads to very low value of `DeltaT_(b)orDeltaT_(f)`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 If molar mass is low,`DeltaT_(b)orDeltAT_(f)`being low cannot be read out accurately .A little error in measurement of `DeltaT_(b)` will cause abnormal values of mol.wt. |
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| 929. |
Assertion : Boiling point of water tis `100^(@)C`although water boils below `100^(@)C` on mountains. Reason:Boiling point of a liquid is th e temperature at which V.P.of liquids become equal to `1`atm.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 Water boils at low temperature at mountains where atmospheric pressure is low ,i.e.,when `P^(@)` =atmospheric pressure. |
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| 930. |
Study the figures given below and mark the correct statement. A. (i)Nitric acid + water, (ii)Acetone + Ethyl alcoholB. (i)Water + Ethyl alcohol , (ii) Acetone + BenzeneC. (i)Acetone + Ethyl alcohol , (ii)Acetone + ChloroformD. (i)Benzene + Chloroform , (ii)Acetone + Chloroform |
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Answer» Correct Answer - C Acetone +ethyl alcohol solution shows positive deviation while acetone + chloroform shows negative deviation. other examples : Positive deviations - Acetone + ethyl alcohol, acetone + benzene, water + ethyl alcohol . Negative deviations - Nitric acid + water, benzene + chloroform |
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| 931. |
For an ideal solution with `P_A gt P_B` , which of the following is true ?A. `(x_A)_"liquid"=(X_A)_"vapour"`B. `(x_A)_"liquid"gt(X_A)_"vapour"`C. `(x_A)_"liquid"lt(X_A)_"vapour"`D. `(x_A)_"liquid"` and `(X_A)_"vapour"` do not bear any relationship with each other |
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Answer» Correct Answer - C As A is more volatile , vapour phase would be richer in A. Hence `(x_A)_"liquid" lt (x_A)_"vapour"` |
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| 932. |
Define the following terms : (a) Ideal solution (b) Azeotrope (c) Osmotic pressure |
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Answer» (a) Ideal solution: Those solutions which follow Raoult's law at specific temperature over the entire range of concentration are called Ideal solution. (b) Azeotrope: A liquid mixture which distills at constant temperature without under going any changes in composition is called Azeotrope. (c) Osmotic pressure: The minimum excess pressure that has to be applied on the solution side to prevent the entry of the solvent into the solution through the semipermeable membrane is called osmotic pressure. |
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| 933. |
(a) Why is the vapour pressure of a solution of glucose in water lower than that of water ? (b) A `6.90M` solution of `KOH` in water contains `30%` by mass of `KOH`. Calculate the density of the `KOH` solution. [Molar mass of `KOH=56g "mol"^(-1)]` |
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Answer» Correct Answer - (a) When glucose is dissolved in water, the surface molecules consist of both glucose and water, therefore, escaping tendency of molecules of solvent in vapour decreases, therefore, vapour pressure decreases. (b) `M=(% "by massxxdxx10)/("Mol wt.") rArr 6.90=(30xxdxx10)/(56)` `rArr d=(6.90xx56)/(30xx10)=(386.4)/(300)=1.288g cm^(-3)` |
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| 934. |
Calculate the mole fraction of `H_(2)SO_(4)` in a solution containing `98% H_(2)SO_(4)` by mass. |
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Answer» Given a solution containing - `98% H_(2)SO_(4)` by mass. It means 98 gms of `H_(2)SO_(4)` and 2 gms of `H_(2)O` mixed to form a solution. `n_(H_(2)O)=("Weight")/("GMW")=(2)/(18)=(1)/(9)` `n_(H_(2)S_(o4)=("Weight")/("GMW")=(98)/(48)=1` Mole fraction of `H_(2)SO_(4)." X"_(H_(2)SO_(4))=(n_(H_(2)SO_(4)))/(n_(H_(2)O)+n_(H_(2)SO_(4)))` `=(1)/(1//9+1)=(9)/(10)=0.9` |
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| 935. |
What is relative lowering of vapour pressure ? |
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Answer» Relating lowering of vapour pressure : The ratio of lowering of vapour pressure of a solution containing non-volatile solute to the vapour pressure of pure solvent is called relative lowering of vapour pressure. `"R.L.V.P."=(P_(0)-P_(S))/(P_(0))" "P_(o)-P_(S)` = lowering of vapour pressure `P_(o)` = Vapour pressure of pure solvent |
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| 936. |
The depression in freezing point of water observed for the same amount of acetic acid, dichloro-acetic acid and trichloro acetic acid increases in the order given above. Explain briefly. |
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Answer» Given acids are `CH_(3)COOH,CHCl_(2)COOH, C Cl_(3)COOH`. The depression in freezing point of a solute in water depends upon the number of ions (or) particles in aqueous solution. The given three acdis are arranged in the order of their acidic strengths (increasing order) `CH_(3)COOH lt CHCl_(2)COOH lt C Cl_(3)COOH` Due to the presence of three chlorine atoms `C Cl_(3)COOH` is more acidic than `CHCl_(2)COOH` and followed by `CH_(3)COOH`. `:.` The order of depression in freezing point is `CH_(3)COOH lt CHCl_(2)COOH lt C Cl_(3)COOH` |
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| 937. |
The vapour pressure of pure benzene `C_(6)H_(6)` at `50^(@)C` is `268 "torr"`. How many moles of non-volatile solute per mole of benzene is required to prepare a solution of benzene having a vapour pressure of `167 "torr"` at `50^(@)C`?A. 0.377B. 0.605C. 0.623D. 0.395 |
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Answer» Correct Answer - B `{:(P_("benzene")^@=268 "torr"),(" "P_(sol)=167 "torr"):}}=chi_B=(268-167)/268=0.377,chi_B=n_B/(n_A+n_B)` `rArr n_B/n_A=0.605` |
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| 938. |
The boiling point of a solution of `0.0150g` of a substance in `15.84g` of ether was found to be `100^(@)C`higher than that of pure ether.What is the molecular weight of the substance [Molecular elevation constant of ether per`100`A. `144.50`B. `143.18`C. `140.28`D. `146.66` |
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Answer» Correct Answer - B `m=(K_(b)xxwxx1000)/(DeltaT_(b)xxW)=143.18` |
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| 939. |
What amount of `CaCl_2` (i=2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at `27^@` C ?A. 3.42 gB. 9.24 gC. 2.834 gD. 1.820 g |
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Answer» Correct Answer - D `pi=iCRT = I n/V RT` `n=(pixxV)/(ixxRxxT) =(0.5xx2)/(2.47xx0.0821xx300)`=0.0164 mol Amount of `CaCl_2=n xx M =0.0164 xx 111` = 1.820 g |
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| 940. |
The correct relationship between the boiling point f very dilute solution of `"AlCl"_(3)("T"_(1)"K") " and" " CaCl"_(2)("T"_(2)"K")` having the same molar concentration isA. `"T"_(1)="T"_(2)`B. `"T"_(1)gt"T"_(2)`C. `"T"_(2)gt"T"_(1)`D. `"T"_(2)le"T"_(1)` |
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Answer» Correct Answer - B `"i"_("AlCl")gt"i"_("CaCl"_(2))` `thereforeDelta"T"_("b")("AlCl"_(3)" solution")gtDelta"T"_("b")("CaCl"_(2)" solution")` `"Hence"" T"_(1)gt"T"_(2)` |
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| 941. |
Determine the amount of `CaCl(i=2.47)` dissolved in `2.5L` of water such that its osmotic pressure is `0.75atm ` at `27^(@)C`. |
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Answer» We know that, `pi = i(n)/(V) RT` `rArr pi =i(w)/(MV) RT rArr w = (piMV)/(iRT)` `pi = 0.75 atm` `V = 2.5 L` `i= 2.47` `T =(27 +273)K = 300 K` Here, `R =0.082L atm K^(-1)mol^(-1)` `M = 1 xx 40 +2 xx 35.5 = 111g mol^(-1)` Therefore, `w = (0.75 xx 111 xx 2.5)/(2.47 xx 0.0821 xx 300) = 3.42g` Hence, the required amount of `CaCI_(2)` is `3.42g`. |
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| 942. |
At `40^(@)C`, vapour pressure in Torr of methanol nd ethanol solution is `P=119x+135` where `x` is the mole fraction of methanol. HenceA. vapour pressure of pure methanol is `119` TorrB. vapour pressure of pure ethanol is `135` TorrC. vapour pressure of equimolar mixture of each is `127` TorrD. mixture is completely immiscible |
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Answer» `P=119x+135` `x=1` for pure methanol. so `P^(@)_("methanol")=119+135=254` Torr But for pure ethanol `x=0` so `P_("ethanol")^(@)=135` Torr |
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| 943. |
Semi-permeable membrane is chemicallyA. Copper ferrocyanideB. Copper ferricyanideC. Copper sulphateD. Potassium ferrocyanide |
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Answer» Correct Answer - A It is `Cu_(2) Fe(CN)_(6)` molecule is gelatinous state of work as semi-permeable in nature. |
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| 944. |
At `323K`, the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation `p=120X_(A)+140`, where `X_(A)` is the mole fraction of methanol. Then the value of `underset(x_(A) to 1)(lim) (p_(A))/(X_(A))` is:A. `250mm`B. `140mm`C. `260mm`D. `20mm` |
| Answer» Correct Answer - C | |
| 945. |
When a substance is dissolved in a solvent the vapour pressure of solvent decreases. This brings:A. an increases in b.pt of the solutionB. a decreases in b.pt of a solutionC. an increases in f.pt of the solventD. none |
| Answer» Correct Answer - A | |
| 946. |
In the diagram given in the point D refers to composition where, the solution exhibitsA. Maximum boiling azeotropyB. Minimum boiling azeotropyC. Highest vapour pressureD. Lowest boiling point |
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Answer» Correct Answer - A If vapour pressure is lower the boiling point increases. |
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| 947. |
A liquid mixture which boils without changes in the composition is called a/anA. binary liquid mixtureB. azeotropic mixtureC. isotropic mixtureD. no specific name |
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Answer» Correct Answer - B It is called an azeotrope (see Comprehensive Review) |
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| 948. |
Depression in freezing point of solution of electrolytes are generallyA. lowerB. higher than what should be normallyC. low or high depending upon nature of electrolyteD. what it should be normally |
| Answer» Correct Answer - C | |
| 949. |
If the vapour pressure of solutions of two liquids are less than those expected ideal solution they are said to have:A. negative deviation from ideal behaviourB. positive deviations from ideal behaviourC. ideal behaviourD. positive deviation for lower concentration and negative deviations for higher concentration |
| Answer» Correct Answer - A | |
| 950. |
If the vapour pressure of solutions of two liquids are less than those expected ideal solution they are said to have:A. negative deviations from ideal behaviourB. positive deviations from ideal behaviourC. ideal behaviourD. positive deviations for lower concentration and negative deviations for higher concentration |
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Answer» Correct Answer - A It is a solution with -ve devations (see Comprehensive Review for details) |
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