Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q

What is the molar solubility of Al\left ( OH \right )_{3} in 0.2M\: \: \: \: NaOH solution? Given that, solubility product of Al\left ( OH \right )_{3}=2.4\times 10^{-24}:

 

 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Solubility and K{H} values - Gases having more values will have less solubility. -         Option 1)Option 2)Option 3)Option 4)

1\; g of a non-volatile non-electrolyte solute is dissolved in 100 \; g of two different solvents A and B whose ebulliscopic constants are in the ratio of 1:5. The ratio of the elevation in their boiling points,\frac{\Delta T_{b}\left ( A \right )}{\Delta T_{b}\left ( B \right )}, is :

  • Option 1)

    5:1

  • Option 2)

    10:1

  • Option 3)

    1:5

  • Option 4)

     1:0.2

                                   given  ------------------------(1)                                                                                   -------------------------(2)                        Option 1)Option 2)Option 3)Option 4) 

Molal depression constant for a solvent is 4.0\:\:K\:kg\:mol^{-1}. The depression in the freezing point of the solvent for 0.03\:\:\:mol\:kg^{-1}\:\:solution\:\:of\:\:K_{2}SO_{4}  is :

( Assume complete dissociation of the electrolyte )

  • Option 1)

    0.18\:K

  • Option 2)

    0.24\:K

  • Option 3)

    0.12\:K

  • Option 4)

    0.36\:K

  Depression in freezing point - -     Mathematical Expression of Depression in Freezing point -   - wherein m = molarity of solvent  = cryoscopic  constant     molal depress const Units =     Vant Hoff factor (i) - In case of electrolytes the observed colligative property is different from theoritical colligative property. There ratio is defined by Vant Haff factor - wherein...

The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 \; M BaCl_{2} in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY\left ( in \; mol\; L\; ^{-1} \right ) in solution is :

  • Option 1)

     4\times 10^{-2}                

  • Option 2)

     6\times 10^{-2}

  • Option 3)

     4\times 10^{-4}

  • Option 4)

    16\times 10^{-4}

 
Given,        Option 1)                   Option 2)   Option 3)   Option 4)

Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 \; mmHg, respectively, at the same temperature. Then correct statement is :

\left ( x_{M}=Mole\; fraction \; of\; 'M' \; in\; solution;

x_{N}=Mole\; fraction \; of\; 'N' \; in\; solution;

y_{M}=Mole\; fraction \; of\; 'M' \; in\; vapour\; phase;

y_{N}=Mole\; fraction \; of\; 'N' \; in\; vapour\; phase )

 

  • Option 1)

      \frac{x_{M}}{x_{N}}=\frac{y_{M}}{y_{N}}       

     

  • Option 2)

    \left ( x_{M}-y_{M} \right )< \left ( x_{N}-y_{N} \right )

  • Option 3)

      \frac{x_{M}}{x_{N}}< \frac{y_{M}}{y_{N}}

  • Option 4)

     \frac{x_{M}}{x_{N}}> \frac{y_{M}}{y_{N}}

 
More volatile component will have greater composition in vapour phase as compared to it's composition in liq.phase The vapour phase of pure liq.  &  are  of  and  of   respectively           ( more volatile) Option 1)             Option 2) Option 3)    Option 4)  

For the solution of the gases, w,x,y and z in water at 298 K, the henry's law constants (KH) are 0.5,2,35 and 40 kbar, respectively. The correct plot for the given data is 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

                                                                                                                    Option 1)Option 2)Option 3)Option 4)

The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture.  The mole fraction of liquids B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are :
 

  • Option 1)

    450\; mmHg,0.5,0.5

  • Option 2)

    450\; mmHg,0.4,0.6

  • Option 3)

    500\; mmHg,0.4,0.6

     

  • Option 4)

    500\; mmHg,0.5,0.5

                                                                      mole fraction of A in vapour phase Option 1)Option 2)Option 3)  Option 4)

The molar solubility of Cd(OH)_{2} is 1.84\times 10^{-5}M in water. The expected solubility of Cd(OH)_{2} in a buffer solution of pH = 12 is :

 

  • Option 1)

    1.84\times 10^{-9}M

     

     

     

  • Option 2)

    \frac{2.49}{1.84}\times 10^{-9}M

  • Option 3)

    6.23\times 10^{-11}M

  • Option 4)

    2.49\times 10^{-10}M

 of                                  S               2S+    2S - > this OH comes from dissociation of      - > this OH comes from a buffer                                        Given                            Option 1)      Option 2)Option 3)Option 4)

A solution is prepared by dissolving 0.6 g of urea (molar mass =60g mol^{-1} ) and 1.8 g of glucose (molar mass = 180 g mol^{-1} ) in 100 mL of water at 27^{\circ}C. The osmotic pressure of the solution is:

(R=0.08206 L atm K^{-1}mol^{-1})

  • Option 1)

    8.2 atm

  • Option 2)

    2.46 atm

  • Option 3)

    4.92 atm

  • Option 4)

    1.64 atm

molar of Urea                                                      Given Water = 100ml molar of glucose =                                                   Temp = 300 k Total moles =  Osmotic pressure =  Option 1)8.2 atmOption 2)2.46 atmOption 3)4.92 atmOption 4)1.64 atm
Dissociation of  : Hence, one molecule sodium sulphate dissociates to form three molecules. Now, Van't Hoff's factor, , where  is degree of dissociation;  Now,  where, C denotes concentration, R denotes universal gas constant and T is temperature in Kelvin.here, Therefore, osmotic pressure is .
7th
15486380425468794424014177925083.jpg Equal volumes of 0.5M of HCL ,0.25M of NaOH and 0.75M of NaCl are mixed. The molarity of NaCl solution is- (A) 0.75M (B) 1/3M(C) 0.50M (C) 2.00M
the molarity of NaCl in the solution should be 1/3 i.e option B should be correct.0.25 M NaOH will act as a limiting reagent for the reaction between HCl and NaOH. Assume volume of each before mixing is 1L                                 HCl      +    NaOH →NaCl + H2O before reaction        0.50*1         0.25*1    0.75*1    --  after reaction           0.25             0        ...
Screenshot_811.png A mixture of 100 mmol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting the solution, respectively, are (Molar mass of Ca(OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6)
Na2SO4 + Ca(OH)2 → CaSO4 + 2 NaOH  mmol of Na2SO4 = 2*1000/143                           =13.98 m Mol mmol of CaSO4 formed = 13.98 m Mol  Mass of CaSO4 formed = 13.98 × 10^-3 × 136 = 1.90 g mmol of NaOH = 28 mmol Ca(OH)2 → Ca2+ + 2 OH- [OH–] = 28 mmol / 100ml=0.028 mol/0.1 L =0.28mol L-1
15482276374931609873616.jpg At 760 Torr pressure and 20 c temperature 1L of water dissolve 0.04 of pure oxygen or 0.02 gm of pure nitrogen. Assuming that dry air is composed of 20 % oxygen and 80% nitrogen (by volume), the masses(in g/L) of oxygen and nitrogen dissolved by 1L of water at 20 c exposed to air at a total pressure 760 torr are:-
Mass of oxygen = 0.04*(20/100) = 0.008g/L Mass of nitrogen=0.02*(80/100) = 0.016g/L
  1. 60 ml \frac{M}{10} HCl + 40 ml \frac{M}{10} NaoH
  2. 55 ml \frac{M}{10} HCl + 45 ml \frac{M}{10} NaOH
  3. 75 ml\frac{M}{5} HCl + 25 ml \frac{M}{5}NaOH
  4. 100 ml \frac{M}{10} HCl + 100 ml \frac{M}{10} NaOH

pH of which one of them will be equal to 1 ?

The p(H) scale - Hydronium ion concentration in molarity is more conveniently expressed on a logarithmic scale known as p(H) scale. - wherein The p(H) of a solution is defined as negative logarithm to base 10 of the activity of hydrogen ion    As N1V1 > N2V2 so acid left at the end of the reaction   So acid is left at the end of the reaction Option 4) c This is correct.

The freezing point of a diluted milk sample is found to be -0.2^{0}C , while it should have been -0.5^{0}C for pure milk. How much water has been added  to pure milk  to make the diluted sample ?

  • Option 1)

    2 cups of water to 3 cups of pure milk.

  • Option 2)

    3 cups of water to 2 cups of pure milk.

  • Option 3)

    1 cup of water to 3 cups of pure milk.

  • Option 4)

    1 cup of water to 2 cups of pure milk.

  Depression in freezing point - -       Freezing -  Freezing  occurs when liquid solvent is in equilibrium with solid solvent. As non volatile solute decreases, the vapour pressure freezing point decreases. -As we knoew that Freezing point of milk Freezing point of milk (diluted)      Option 1)2 cups of water to 3 cups of pure milk.Option 2)3 cups of water to 2 cups of pure milk.Option 3)1...

Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :

  • Option 1)

    C : liquid in solid ; M : Liquid in Solid ; S solid in gas 

  • Option 2)

    C : liquid in solid ; M : Liquid in liquid ; S : solid in gas 

  • Option 3)

    C : solid in liquid  ; M : Liquid in Liquid ; S : solid in gas 

  • Option 4)

    C : solid in liquid  ; M : Solid in liquid; S : solid in gas 

  Types of solutions - a)    Liquid solute and Liquid solvent b)    Solid solute and Liquid solvent c)    Gaseous solute and Liquid solvent d)    Solid solute and Solid solvent e)    Liquid solute and Solid solvent f )    Gaseous solute and Solid solvent g)    Solid solute and Gaseous solvent h)    Liquid solute and Gaseous solvent i)     Gaseous solute and Gaseous solvent -       Solutions - A...

K_{2}HgI_{4} is 40 % ionised in aqueous solution. 

The value of its van't Hoff factor (i) is :

  • Option 1)

    2.2

  • Option 2)

    1.8

  • Option 3)

    2.0

  • Option 4)

    1.6

  Vant Hoff factor (i) - In case of electrolytes the observed colligative property is different from theoritical colligative property. There ratio is defined by Vant Haff factor - wherein       Vant Hoff factor for dissociation - Where is the no. of dissociated particles degree of dissociation   - wherein       Vant Hoff factor for Association - number of particles associated ...

If K_{sp} of Ag_{2}CO_{3} is 8 \times 10^{-12}, the molar solubility of Ag_{2}CO_{3} in 0.1 M AgNO_{3} is :

  • Option 1)

    8 \times 10^{-11} M

     

  • Option 2)

    8 \times 10^{-12}

  • Option 3)

    8 \times 10^{-10}

  • Option 4)

    8 \times 10^{-13} M

  Relation between concentration terms (Molality & Mole fraction) - - wherein         Relation between concentration terms (Molality & Molarity) - - wherein As we have learned in solution The reaction is      Option 1) M  Option 2) M Option 3) M Option 4) M

Elevation in the boiling point for 1 molal solution og glucose is 2 K .the depression inthe freezing point for 2molal solution of glucose in the same solvent is 2  ,K.the relatin between Kb and Kfis:

  • Option 1)

    Kb = Kf

  • Option 2)

    Kb =1.5 Kf

  • Option 3)

    Kb =0.5 Kf

  • Option 4)

    Kb =2 Kf

  Mathematical Expression - Unis of   - wherein     Mathematical Expression of Depression in Freezing point -   - wherein m = molarity of solvent  = cryoscopic  constant     molal depress const Units =   As we have learned The change in Tb and Tf . Option 1)Kb = KfOption 2)Kb =1.5 KfOption 3)Kb =0.5 KfOption 4)Kb =2 Kf

A mixture of 100 m mol of Ca(OH)_{2} and 2 g of sodium sulphate was dissolved in water and the volume was made up to 100 mL .The mass of calcium sulphate formed and the concentration of OH^{-} in resulting solution , respectively , are  : (Molar mass of  Ca(OH)_{2}  ,Na_{2}SO_{4} \: and \: \: CaSO_{4} are 74, 143 and 136 g mol^{-1} , respectively ; K_{sp} \: \: of\: \: Ca(OH)_{2}\: \: is\: \: 5.5\times 10^{-6})

  • Option 1)

    1.9\: g,\: \: 0.28\: mol\: L^{-1}

  • Option 2)

    13.6g,\: \: 0.28\: molL^{-1}

  • Option 3)

    1.9g,\: \: 0.14molL^{-1}

  • Option 4)

    13.6g,\: \: 0.14molL^{-1}

@    Mole Fraction - - As we have learned in mole concept . The reaction is -       Vapour Pressure - It is defined as pressure exerted by vapours on liquid surface at equilibrium and condensation. -        Raoult's Law - The partial pressure of any volatile constituents of a solution at a given temperature is equal to the product of vapour pressure of pure constituent and its mole fraction...
Exams
Articles
Questions