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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.
| 1101. |
`3.24gHg(NO_(3))_(2)`(molecules mass`324`)dissolved in `1000` g of water constitutes a solution having a freezing point `-0.0558^(@)C` while `21.68g` of `HgCl_(2)` (Molecular mass`271` in `2000g` of water constitutes a solution with a freezing point of `-0.0744^(@)C`.The `K_(f)` of water is `1.86K Kg mol^(1)`.About the state of ionisation of the two solids in water can be inferred that.A. `Hg(NO_(3))_(2)`is fully ionised but `HgCl_(2)`is fully unionisedB. `Hg(NO_(3))_(2) and HgCl_(2)`both are completely ionisedC. `Hg(NO_(3))_(2)`is fully unionised but `HgCl_(2)`is fully ionisedD. Both`Hg(NO_(3))_(2)and HgCl_(2)` are completely unionised |
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Answer» Correct Answer - A `0.0558=ixx(3.24)/(324)xx1.86` `i=3(100% "ionised")` `0.0744=ixx(21.68)/(271)xx1.86` `i=1 ("unionised")` |
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| 1102. |
5 % solution of glucose is isotonic with a solution of urea (M = 60). Hence the weight of urea present in the solution is :(a) 1.67 g (b) 6.0 g (c) 18.6 g (d) 1.2 g |
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Answer» Option : (a) 1.67 g |
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| 1103. |
Anhydrous sodium sulphate dissolves in water with the evolution of heat. What is the effect of temperature on its solubility ? |
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Answer» Since the dissolution of anhydrous sodium sulphate in water is an exothermic process due to evolution of heat, according to Le Chatelier’s principle its solubility decreases with the increase in temperature. |
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| 1104. |
A solution was prepared by dissolving `6.0 g` an organic compound in `100 g` of water. Calculate the osmotic pressure of this solution at `298 K`,when the boiling point of the Solution is `100.2^(@)C`. (`K_(b)` for water =`0.52 K m^(-1)`), R=0.082 L atm `K^(_1) mol^(-1)`) |
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Answer» The given values are `W_("solute")=6.0 g`,`W_("solvent")=100 g`,`K_(b)=0.52 K m^(-1)` `DeltaT_(b)=100.2-100=0.2` Now using the formula `Mw_("solute")=(K_(b)xxW_("solute")xx1000)/(DeltaT_(b)xxW_("solvent"))` `=(0.52xx6.0xx1000)/(0.2xx100)` `=156 g mol^(-1)` Now, `pi=(nRT)/(V)`,`n=W_("solute")/(Mw_("solute"))` `:.n=6.0/156` `:. n=(6.0xx0.082xx298)/(156xx0.1)` `=9.398 atm` Thus, the osmotic pressure of the solution is `9.398 atm`. |
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| 1105. |
Osmotic pressure of a solution containing 2 g dissolved protain per `300cm^(3)`id solution is 20 mm of Hg at `27^(@)C`. Calculate the molecular mass of protein. (R=0.082 L atm `K^(-1)mol^(-1)`) |
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Answer» Correct Answer - 6238.6 h `mol^(-1)` `W_(B)=2.0g, V=300 cm^(3)=300 mL=0.3 L` `pi=20" mm of Hg"=20/760=0.02632" atm, "R=0.0821" L atm K"^(-1)mol^(-1)` `T=27^(@)C=27+273=300K, M_(B)=?` `M_(B)=(W_(B)xxRxxT)/(pixxV)=((2.0g)xx(0.0821"L atm K"^(-1)mol^(-1))xx(300K))/((0.02632 atm)xx(0.3 L))=6238.6"g mol"^(-1)`. |
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| 1106. |
100 mg of a protein was disoved in just enough water to make 10 mL of the solution. If the solution has an osmotic pressure of 13.3 mm Hg at `25^(@)C`, what is the mass of prtein `(R=0.0821 L atm mol^(-1)K^(-1))` |
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Answer» Osmotic pressure `(pi)=(W_(B)xxRxxT)/(M_(B)xxV)` or `M_(B)=(W_(B)xxRxxT)/(pixxV)` `pi=13.3 , Hg=13.3mm Hg=(13.3)/(760) atm, V=10 mL=10/1000=0.01L` `W_(B)=100mg=100/1000=0.1g, T=25^(@)C=25+273=298K` `R=0.0821 L atm mol^(-1)K^(-1)` `M_(B)=((0.1g)xx0.0821(atm K^(-1)Mol^(-1))xx298K)/(((13.3)/760atm)xx(0.01 L))` =`13980.45 g mol^(-1)=1.45xx10^(4)g mol^(-1)` |
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| 1107. |
Osmotic pressure of a solution containing 7 g of a protein per 100 mL of solution is 25 mm Hg at 310 K. Calculate the mass of the protein. (R=0.0821L atm `K^(-1)mol^(-1)`) |
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Answer» Correct Answer - 5159.29 g `mol^(-1)` `W_(B)=7g, V=100mL=0.1 L, pi=25" mm of Hg"=25/760=0.032895 atm` `T=310 K, R=0.0821" L atm K"^(-1)mol^(-1), M_(B)=?` ` M_(B)=(W_(B)xxRxxT)/(pixxV)=((7g)xx(0.021"L atm K"^(-1)mol^(-1))xx(310 K))/((0.032895" atm")xx(0.1L))=54159.29" g mol"^(-1)` |
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| 1108. |
Identify the correct statement (a) vapour pressure of solution is higher than that of pure solvent. (b) boiling point of solvent is lower than that of solution (c) osmotic pressure of solution is lower than that of solvent (d) osmosis is a colligative property. |
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Answer» Correct answer is (b) boiling point of solvent is lower than that of solution. |
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| 1109. |
For a solution, the plot of osmotic pressure (π) verses the concentration (e in mol L-1 ) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is …(a) 310 x 0.082 K (b) 3 10°C (c) 37°C(d) 310 / 20.082 |
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Answer» (c) 37°C π = CRT y = x(m) m = RT 310 R = RT T = 310 K = 37°C |
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| 1110. |
`PtCl_(4).6H_(2)O` can exist as a hydrated complex. `1 m` aqueous solution has the depression in freezing point of `3.72^(@)`. Assume `100%` ionization and `K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg`, then the complex isA. `[Pt(H_(2)O)_(6)]Cl_(4)`B. `[Pt(H_(2)O)_(4)Cl_(2)]Cl_(2).2H_(2)O`C. `[Pt(H_(2)O)_(3)Cl_(3)]Cl_(3).3H_(2)O`D. `[Pt(H_(2)O)_(2)Cl_(4)].4H_(2)O` |
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Answer» Correct Answer - C `DeltaT_(f)="Molality" xx K_(f) xx i` `i=2`, hence complex is binary. Also, CN of `Pt` is `6` and complex is hydrated. |
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| 1111. |
Which of the following is dependent on temperature?A. MolarityB. Mole fractionC. Weight percentageD. Molality |
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Answer» Correct Answer - A Temperature dependent unit is molarity. |
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| 1112. |
What will be the molarity of a solution containing `5`g of sodium hydroxide in`250ml` solution?A. `0.5`B. `1.0`C. `2.0`D. `0.1` |
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Answer» Correct Answer - A `M=(W)/(m.wt.)xx(1000)/("Volume in ml")` `=(5xx1000)/(40xx250)=0.5M` |
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| 1113. |
How many moles of sucrose should be dissolved in `500gms` of water so as to get a solution which has a difference of `104^(@)C` between boiling point and freezing point. `(K_(f)=1.86 K Kg "mol"^(-1).K_(b)=0.52K Kg "mol"^(-1))`A. `1.68`B. `3.36`C. `8.40`D. `0.840` |
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Answer» Correct Answer - D Boiling point of solution `=` boiling point `+DeltaT_(b)=100+DeltaT_(b)` Freezing point of solution `=` Freezing point `-DeltaT_(f)=0-DeltaT_(f)` Diffrence in temperature (given) `=100+DeltaT_(B)-(-DeltaT_(f))` `104=100+DeltaT_(b)+DeltaT_(f)=100+"molality"xxK_(b)+"molallity"xxK_(f)` `=100+"molality"(0.52+1.86)` `therefore "Molality"=(104-100)/(2.38)=(4)/(2.38)=1.68m` and molality `=("moles"xx1000)/(W_(gm(solvent))),1.68=("moles"xx1000)/(500)` `therefore` Moles of solute `=(1.68xx500)/(1000)=0.84` moles. |
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| 1114. |
A `20.0mL` sample of `CuSO_(4)` solution was evaporated to dryness, leaving `0.967g` of residue. What was the molarity of the original solution ? `(Cu=63.5)`A. `48.4M`B. `0.0207M`C. `0.0484M`D. `0.303M` |
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Answer» Correct Answer - D Molarity `=(0.967//159.5)/(20//1000) {"Mol" wt. of underset(159.5)(CuSO_(4))}` `=0.303M` |
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| 1115. |
Consider which can be expressed is degree (temperature) areA. III,IVB. I,IIC. I,II,IIID. I,III |
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Answer» Correct Answer - C As `DeltaT_(b)=iK_(b)m` so `iK_(b)m` can be expressed in degree (Unit of temperature) and `K_(b)m` can be expressed in degree (Unit of temperature) and `(DeltaT_(b))/(i)` can be expressed in degree (Unit orf temperature) But unit of `K_(b)` is `"mol"^(-1)kg K` |
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| 1116. |
Colligative properties have many pracitical uses, some of them may be: I: Melting of snow by slat II: Desalination of sea water III: Desalination of sea water III: Determination of molar mass IV: Determination of melting point and boiling point of solvent Actual practical uses are :A. I,IIB. III,IVC. I,II,IIID. II,III,IV |
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Answer» Correct Answer - C I. Melting of snow by salt: Depression in freezing point II. Desalination of sea water: Reverse osmosis III. Osmosis is used to determine the molar mass. |
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| 1117. |
Select correct statement(s):A. When solid `CaCl_(2)` is added to liquid water, the boiling temperature risesB. When solid `CaCl_(2)` is added to ice at `0^(@)C`, the freezing temperature fallsC. Both (A) and (B)D. None of the above |
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Answer» Correct Answer - C When non volatile solute added to solvent. Due to elevation in boiling point, boiling point `uparrow` and due to dispression in freezing point, freezing temperature `downarrow` |
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| 1118. |
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g `"mol"^(-1)`), vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance ? |
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Answer» The various quantities known to us are as follows. `P_(1)^(0)=0.850" bar"` `P=0.845"bar"` `M_(1)=78" g mol"^(-1)` `w_(2)=0.5" g"` `w_(1)=39"g"` Substituting these values in equation `(P^(0)-P)/(P_(1)^(0))=(w_(2)xxM_(1))/(M_(2)xx w_(1))` we get `(0.850" bar"-0.845" bar")/(0.850" bar")=(0.5"g"xx78" g mol"^(-1))/(M_(2)xx39" g")` Therefore, `M_(2)=170" g mol"^(-1)`. |
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| 1119. |
The vapour pressure of pure benzene at a certain temperature is `0.850` bar. A non-volatile, non-electrolyte solid weighting `0.5 g` when added to `39.0 g` of benzene (molar mass `78 g mol^(-1)`). The vapour pressure of the solution then is `0.845` bar. What is the molar mass of the solid substance?A. 170 amuB. 160 amuC. 120 amuD. 190 amu |
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Answer» Correct Answer - 1 we are given `P_(1)^(0)=0.850` bar, `P_(1)`=0.845 bar `M_(1)=78 g mol^(-1), W_(1)=39 g and W_(2)=0.5 g` Substituting these values in Equation `(2.58)`, we get `DeltaP_(1)//P_(1)^(0)=(W_(2)xxM_(1))/(W_(1)xxM_(2))` `((0.850 - 0.845) "bar")(0.850 "bar")=(0.5 gxx78 g mol^(-1))/(M_(2)xx39 g)` or `M_(2)=170 g mol^(-1)` Thus, molecular mass of noncolatile solid is 170 amu |
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| 1120. |
Which one of the following is not a colligative property?A. Osmotic pressureB. Critical temperatureC. Elevation in boiling point of solventD. Depression of freezing point of solvent |
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Answer» Correct Answer - 2 There are four colligative properties of a dilutesolution containing a nonvolatile solute in a volatile solvent that are directly proportional to the number of solution particles present. They are (1) vapor pressure lowering (or relativelowering of vapor pressure) (2) boiling point elevation, (3) freezing point depression and (4) osmotic pressure |
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| 1121. |
The normal boiling point of toluene is `110.7^(@)C` and its boiling point elevation constant `3.32 " K kg mol"^(-1)`. The enthalpy of vaporization of toluene is nearly :A. `"17 kJ mol"^(-1)`B. `"21 kJ mol"^(-1)`C. `"51 kJ mol"^(-1)`D. `"68 kJ mol"^(-1)` |
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Answer» Correct Answer - B `K_(b)=(RT_(0)^(2))/(1000L),3.32=((8.314xx10^(-3))xx(383.7)^(2))/(1000xxL)` `L=0.368kJ//g` Latent heat per mol `= 0.368 xx` molar mass of acetone `= 0.368xx58=21.344" kJ mol"^(-1)`]. |
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| 1122. |
Assertion: One molar aqueous solution is more concentrated than that of 1 molal aqueous solution . Reason : Molarity is a function of temperature as volume depends on temperature.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is falseD. If both assertion and reason are false |
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Answer» Correct Answer - B 1 molar solution has 1 mole of solute per litre of solution or per `1000 cm^3` of solution. Now , density of water = 1g/cc `therefore 1000 cm^3`=1000g Hence, 1 mole is present in less than 1000 g of solvent . |
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| 1123. |
`PtCl_(4).6H_(2)O` can exist as a hydrated complex. `1 m` aqueous solution has the depression in freezing point of `3.72^(@)`. Assume `100%` ionization and `K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg`, then the complex isA. `[Pt(H_(2)O)_(6)]Cl_(4)`B. `[Pt(H_(2)O)_(4)Cl_(3)Cl_(2).2H_(2)O`C. `[Pt(H_(2)O)_(3)]Cl.3H_(2)O`D. `[Pt(H_(2)O)_(2)Cl_(4)].4H_(2)O` |
| Answer» Correct Answer - C | |
| 1124. |
The relative decreases in the vapour pressure of an aqueous solution containing `2 mol [Cu(NH_(3)_(3)Cl]` in `3 mol H_(2)O` is `0.50`. On reaction with `AgNO_(3)`, this solution will formA. `1 mol AgCl`B. `0.25 mol AgCl`C. `2 mol AgCl`D. `0.40 mol AgCl` |
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Answer» Correct Answer - A `(DeltaP)/P^(@)=0.50 = (n_(1)i)/(n_(1)i + n_(2))` `0.50 =(2i)/(2i + 3)` `i=1.5 = 1 + x` `x=0.5` Hence,2 mol NaCl will exists as `1 "mol" Cl^(ө)` due to `50%` ionization. |
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| 1125. |
The most likely of the following mixtures to be an ideal solution isA. `NaCl-H_(2)O`B. `C_(2)H_(5)OH-C_(6)H_(6)`C. `C_(6)H_(16)(l)-H_(2)O`D. `C_(6)H_(5)OH-H_(2)O` |
| Answer» Correct Answer - A | |
| 1126. |
Which has the maximum osmotic pressure at temperature `T`?A. `100 mL` of `1 M` urea solution.B. `300 mL` of `1 M` glucose solution.C. Mixture of `100 mL` of `1 M` urea solution and `300 mL` of `1 M` glucose solution.D. All are isotonic. |
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Answer» Correct Answer - D `pi=MRT` `pi=RT` `pi=RT` `pi=RT`(mix has concentration = `1 M`) |
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| 1127. |
The value of `K_(f)`for water is `1.86^(@)`, calculated from glucose solution, The value of `K_(f)`for water calculated for NaCl solution will be,A. `=1.86`B. `lt1.86`C. `gt1.86`D. Zero |
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Answer» Correct Answer - A `K_(b)` is characteristic constant for given solvent. |
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| 1128. |
Mole fraction of component `A` in vapour phase is `chi_(1)` and that of component `A` in liquid mixture is `chi_2`, then (`p_(A)^@`)= vapour pressure of pure A, `p_(B)^@` = vapour pressure of pure B), the total vapour pressure of liquid mixture isA. `(P_(A)^(@)chi_(2))/chi_(1)`B. `(P_(A)^(@)chi_(1))/chi_(2)`C. `(P_(A)^(@)chi_(1))/chi_(2)`D. `(P_(B)^(@)chi_(2))/chi_(1)` |
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Answer» Correct Answer - A `chi_(1)=(P_(A)^(@)chi_(A))/(P_(Total))=(P_(A)^(@)chi_(2))/(P_(Total))` `:. (P_(Total)=(P_(A)^(@)chi_(2))/chi_(1)` |
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| 1129. |
At `25^(@)C`, the vapour pressure of pure methyl alcohol is 92.0 torr. Mol fraction of `CH_(3)OH` in a solution in which vapour pressure of `CH_(3)OH` is 23.0 torr at `25^(@)C`, is:A. 0.25B. 0.75C. 0.5 0D. 0.66 |
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Answer» Correct Answer - A `{:(P_(CH_3OH)=92 "torr"),(" "P_(sol)=23 "torr"):}}=chi_B=(92-23)/92=0.75,chi_A=0.25` `chi_A=0.25` |
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| 1130. |
Which of the following is a colligative property ?A. Lowering of vapour pressureB. Osmotic pressureC. Freezing pointD. Boiling point |
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Answer» Correct Answer - 2 |
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| 1131. |
Deep – sea divers use air diluted with helium gas in their tanks. Why? (or) Justify this statement. |
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Answer» 1. Deep-sea divers carry a compressed air tank for breathing at high pressure under water. This air tank contains nitrogen and oxygen which are not very soluble in blood and other body fluids at normal pressure. 2. As the pressure at the depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathes from tank. 3. When the divers ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood quickly forming bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends” which are painful and dangerous to life. 4. To avoid such dangerous condition they use air diluted with helium gas (11.7 % helium, 56.2% nitrogen and 32.1% oxygen) of lower solubility of helium in the blood than nitrogen. |
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| 1132. |
What are the units of mole fraction ? |
| Answer» Mola freaction being a ratio has no units. | |
| 1133. |
Why does not molality of the solution cange with temperature ? |
| Answer» It takes into account only the mass of the solute as well of the solvent. Both of them do not change with change in temperature. | |
| 1134. |
Calculate the normality of `NaOH` when `2 g` is present in `800 mL` solution. |
| Answer» Correct Answer - `0.04 g//mL` | |
| 1135. |
The vepour pressure of ethanol and mthanol are 44.5 mm and 88.7 mm of Hg respectively. A solution prepared by mixing 60 g of ethanol and 40 g of methanol. Assuming the solution to be idea, calculate the vepour pressure of the mixing 60 g of ethanol and 40 g of methanol. Assming to be ideal, calculate the vepour pressure of the solution. |
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Answer» `"No. of moles of methanol "(n_(B))=("Mass of" CH_(3)OH)/("Gram molar mass")=((40 g))/((32 gmol ^(-1)))=1.25 mol` `" No. of moles of ethanol "(n_(A))=("Mass of" C_(2)H_(5)OH)/("Gram molar mass")=((60g))/(46gmol^(-1))=1.30 mol ` `" Mole fraction of methanol"(x_(B))=n_(B)/(n_(B)+n_(A))=((1.30 mol))/((1.25 mol+1.30 mol))=0.51 ` `"Vapour pressure of pure methanol" (P_(B)^(@))=88.7 mm` `" Vapour pressure of pure ethanol "(P_(A)^(@))=44.5 mm` `"Partial vapour pressure of methanol" (P_(B)^(@))=P_(B)^(@)x_(B)=(88.7mmxx0.49)=43.46 mm` `"Partial vapour pressure of methanol "(P_(A)^(@))=P_(A)^(@)x_(A)=(44.5mmxx0.51)=22.70 mm` since the solution is ideal in nature, `" Total vepur pressure of solution" (P)=P_(A)+P_(B)=(22.70+43.46)=66.166mm` |
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| 1136. |
The vapour pressure of ethly acetate and ethly propionate are72.8mm abd 27.7mm of Hg respectively. A solution is prepared by mixing 25 g ethly acetate and 50 g of ethly propionate. Assuming to be ideal, calculate its vapour pressure. |
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Answer» No. of moles of `CH_(3)COOC_(2)h_(5)(n_(B))=("Mass of" CH_(3)COOC_(2)h_(5))/("Molar mass")` `((25g))/((88g mol^(-1))=0.284 mol` No. of moles of `C_(2)H_(5)COOC_(2)H_(5) (n_(A))=("Mass of" C_(2)H_(5)COOC_(2)H_(5))/("Molar mass")` `=((50g))/((102g mol^(-1))=0.49 mol` Mjole fraction of ethyl acetate `(x_(B)) =n_(B)/((n_(A)+n_(B))=((0.284 mol))/((0.28l4mol+0.49 mol))=0.37 ` Mole fraction of ethyl propionate `(x_(A))=n_(A)/(n_(A)+n_(B))=((0.49 mol))/((0.284 mol+0.49))=0.63` Vapour pressure of ethly acetate `(P_(B)^(@))=72.8 mm` Vapour pressure of ethly propionate `(P_(A)^(@))=27.7 mm ` Vapour pressure of ethyl acetate in solution `(P_(B))=P_(A)^(@)x_(A)=(72.8mm)xx0.37=26.94 mm` Vapour pressure of ethyl propionate in solution `(P_(A))==P_(A)^(@)X_(a)=(27.7 mm)xx0.63=12.45 mm` Total vapour pressure of solution =`P_(B)+P_(A)` =26.94mm+17.45mm=44.39 mm |
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| 1137. |
Methanol and ethanol from nearly an ideal solution. If the vapour pressure of pure methanol and ethanol at `350 K are 8.1xx10^(4)and 4.5xx10^(4)Nm^(-2)` respeclate the total vapour pressure of the solution and the mole fraction of methanol in the vapour phase if the solution contains 64 g of methanol and 46 g of ethanol at this temperature. |
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Answer» Moles of etanol `(n_(B)`=`(Mass of ethanol)/(Molar mass of ethanol(C_(2)H_(5)OH))=((46g))/((46gmol^(-1)))=1 mol` Moles of methanol `(n_(2))`=`(Mass of methanol)/(Molar mass of methanol(CH_(3)OH))=1/3` Mole fraction of methanol `(x_(B))`=`n_(A)/(b_(A)+n_(B))=((2 mol))/((1 moil+ 2mol))=2/3` Mole fractin of methanol (x_(A))= `n_(A)/(n_(A)+n_(B))=((2 mol))/((1 mol + 2mol))=2/3` `"Vapour pressure of ethanol in mixtue "(P_(B))=P_(B)^(@)xx x_(B)=(4.5xx10^(4)Nm^(-2))xx1/3` =`1.5xx10^(4) Nm^(-2)` `"Vapour pressure of methanol in mixture "((P_(A))=P_(A)^(@)xx x_(A))=(8.1xx10^(4) Nm^(-2))xx 2/3` =`5.4xx10^(4) Nm^(-2)` `"Total vepour pressure of mixture"=(1.5xx10^(4)+5.4xx10^(4))=6.9xx10^(4)Nm^(-2)` `"Mole fraction of methanol in the vepour phase" (x_(CH_(3)OH))=P_(A)/(P_(A)+P_(B))=((5.4xx10^(4)Nm^(-2)))/((6.9xx10^(4)Nm^(-2)))` =0.7826 |
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| 1138. |
The vapoure pressure of benzene and toluene at `20^(circ)C` are `75 mm` of `Hg` and `22mm` of `Hg` respectively. `23.4g` of benzene and `64.4g` of toluene are mixed. If two forms ideal solution, calculate the mole fraction of benzene in vapour phase when vapour are in equilibrium with liquid mixture. |
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Answer» Calculation of the total vapour pessure `"Moles of benzene "(n_(B))=("Mass of" CH_(3)COOC_(2)H_(5))/("Molar massof benzene")(C_(6)H_(6))=((23.4g))/((78g mol^(-1)))=0.3` mol `" Mole fraction of benzene"(x_(A))=("Mass of toluene")/("Molar mass jof toluene"(C_(7)H_(8)))= ((64.4g))/((92g mol^(-1)))=0.7` mol `"Mole faction of benzene"(x_(B))=n_(B)/(n_(a)+n_(B))=((0.3mol))/((0.3mol+0.7 mol))=0.3` `"mole fraction of toluene "(x_(A))=n_(A)/(n_(A)+n_(B))/((0.7 mol))/((0.3 mol+0.7mol))=0.7 ` `"Mole fraction of toluene"((x_(A))=n_(A)/((0.3mol+0.7mol))=0.7` `" Vopour pressre of benzene in soulution "(P_(B))=P_(B)^(@)xxx_(B)=(67mm)xx0.3=22.5mm` Calculoation of mnole fractin of benxene in the vapur phase. `"Mole fration of benzene in vapur phase"=("Partial lllvapour prassure of benzene")/("Total vapour pressure")` `((22.5mm))/((37.9mm))=0.59`. |
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| 1139. |
The vapour pressure of pure benzene at `88^(@)C` is `957 mm` and that of toluene at the same temperature is `379.5 mm`. The composition of benzene-toluene misture boiling at `88^(@)C` will beA. `chi_("benzene")=0.66 , chi_("toluene") = 0.34`B. `chi_("benzene")=0.34, chi_("toluene") = 0.66`C. `chi_("benzene")= chi_("toluene") = 0.5`D. `chi_("benzene")=0.75, chi_("toluene") = 0.25` |
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Answer» Correct Answer - a `p = p_("benzene").^(@)chi_("benzene")+ p_("toluence").^(@)chi_("toluence")` 760= 957 |
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| 1140. |
The vapour pressure of pure benzene and pure toluene at a particular temperature are 100 mm and 50 mm repectively. The mole fration of benzene in the vapour phase in contact with equimolar solution os benzene and toluene is:A. 0.67B. 0.75C. 0.33D. 0.5 |
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Answer» Correct Answer - a For equimole solutions, `X_(B)=X_(f)=0.5` `P_(B)=X_(B)xxP_(B)^(@)=0.5xx100=500mm` `P_(T)=X_(T)xxP_(T)=0.5xx50-25 mm` `"Total vapour pressure "(P_("Total")=75mm` Mole fraction of benzene in the vapour phase= `P_(B)/_("Total")=50/75=2/3=0.67.` |
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| 1141. |
The vapour pressure of benzene and pure toluene at `70^(@)C` are 500 mm and 200 mm Hg respectively. If they form an ideal solution what is the mole fraction of benzene in a mixture boiling at `70^(@)C` at a total pressure of 380 mm Hg?A. 0.2B. 0.4C. 0.6D. 0.8 |
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Answer» Correct Answer - C Let mol fraction of Benzene is x `:. 380 = x xx 500 +(1-x) 200` or `x = 0.6` |
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| 1142. |
Following questions are based on the following activites `(A)` with observatons `(O)` and results or reason ( R) . A. If `A` and `O` are correct and `R` is incorrect, markB. If `A` and `O` are correct and `R` is correct, markC. If `A`,`O`, and `R` are all correct, markD. If `A` is correct, and `O` and `R` are incorrect, mark |
| Answer» Correct Answer - C | |
| 1143. |
Following questions are based on the following activites `(A)` with observatons `(O)` and results or reason ( R) . A. If `A` and `O` are correct and `R` is incorrect, markB. If `A` and `O` are correct and `R` is correct, markC. If `A`,`O`, and `R` are all correct, markD. If `A` is correct, and `O` and `R` are incorrect, mark |
| Answer» Correct Answer - A | |
| 1144. |
The osomotic pressure `pi` depends on the molar concentration of the solution `(pi=CRT)`. If two solutions are of equal solute concentration and, hence, have the same omotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypotonic. Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions: Osmotic rise of a solution depends onA. DensityB. TemperatureC. Nature of solventD. All of these |
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Answer» Correct Answer - A::B `pi=CRT` |
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| 1145. |
The osomotic pressure `pi` depends on the molar concentration of the solution `(pi=CRT)`. If two solutions are of equal solute concentration and, hence, have the same omotic pressure, they are said to be isotonic. If two solutions are of unequal osmotic pressures, the more concentrated solution is said to be hypertonic and the more diluted solution is described as hypotonic. Osmosis is the major mechanism for transporting water upward in the plants. Answer the following questions: Isotonic solutions have sameA. DensityB. MolarityC. Osomotic pressureD. Normality |
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Answer» Correct Answer - D Isontonic solutions have some osmotic pressure as well as same concentration, i.e., molarity. `pi prop C` |
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| 1146. |
Following questions are based on the following activites `(A)` with observatons `(O)` and results or reason ( R) . A. If `A` and `O` are correct and `R` is incorrect, markB. If `A` and `O` are correct and `R` is correct, markC. If `A`,`O`, and `R` are all correct, markD. If `A` is correct, and `O` and `R` are incorrect, mark |
| Answer» Correct Answer - D | |
| 1147. |
When `KCl` dissolves in water (assume endothermic dissolution), then:A. `DeltaH=+ve,DeltaS=+ve,DeltaG=+ve`B. `DeltaH=+ve,DeltaS=-ve,DeltaG=-ve`C. `DeltaH=+ve,DeltaS=+ve,DeltaG=-ve`D. `DeltaH=-ve,DeltaS=-ve,DeltaG=+ve` |
| Answer» Correct Answer - C | |
| 1148. |
The dissolution of ammonium chloride in water is an endothermic process but still it dissolves in water readily. Why ? |
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Answer» This is because of entropy change. In this case, `DeltaS` is positive. `NH_(4)Cl (aq) rarr NH_(4)^ (o+) (aq) + Cl^(ө) (aq)` The ions that were held together in the crystalline solid are free and move in all possible directions. Its entropy increases and then `T DeltaS gt DeltaH`. |
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| 1149. |
The vapour pressure of ethanol and methanol are `44.0 mm` 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. |
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Answer» Mol. Mass of ethyl alcolhol `(C_(2)H_(5)OH) = 46` Number of moles of ethylalcohol `= (60)/(46) = 1.304` Mol. Mass of methyl alcohol `(CH_(3)OH)=32` Number of moles of methyl alcohol `= (40)/(32) = 1.25` Mole fraction of ethyl alcohol , `chi_(A)=(1.304)/(1.304+1.25)` `=0.5107 mm Hg` Mole fraction of methyl alcohol ,`chi_(B)=(1.25)/(1.304+1.25)` `=0.4893 mm Hg` Partial pressure of ethyl alcohol `= chi_(A)P_(A)^(@)` `=22.47 mm Hg` Partial pressure of ethyl alcohol `chi_(B)P_(B)^(@)` `= 43.05 mm Hg` Total vapour pressure of solution `= 22.47 + 43.05` `65.52 mm Hg` Mole fraction of methyl alcohol in the vapour `=("Partial pressure of" CH_(3)OH)/("Total vapour pressure" ) =(43.05)/(65.52) =0.6570 ` |
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| 1150. |
A maxima or minima is obtained in the temperature. The composition curve of a mixture of two liquids does not indicateA. That the liquids are immiscible with one anotherB. That the liquids are partially miscible at the maximum or minimum.C. An azeotropic mixture.D. A eutectic formation. |
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Answer» Correct Answer - A::B::D Greater is the intermolecular force on mixing, more negative will be the devaition `p lt P_(A)^(@)chi_(A) + p lt P_(B)^(@)chi_(B)`. Experimental vapour pressure will be less than calculated vapour pressure. |
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