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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 .
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 OHâ€“ in resulting the solution, respectively, are (Molar mass of Ca(OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g molâ€“1, respectively; Ksp of Ca(OH)2 is 5.5 Ã— 10â€“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...
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