Solutions(12th Grade > Chemistry ) Questions and Answers
Explanation:-
Answer: Option B. -> In a solution showing negative deviation from Raoult's law A-B type of molecular interactions are stronger than A-A and B-B type of molecular interactions.:
B
a)△Hmix>0,△Vmix>0 aretrueformixturesshowing positivedeviations from Raoult'slaw only.Hence option (a)isincorrect
b) It is true
c) Nitric acid and water forms intermolecular H-bonding. Hence shows Negative deviation from Raoult's law.
d) Azeotropes are of two types – minimum and maximum boiling azeotropes. Only minimum boiling Azeotrope boils at temperature below the boiling temperature of its constituents. Hence option (d) is incorrect.
Explanation:-
Answer: Option C. -> Solubility of gases in water decreases as the temperature increases due to increased Henry’s constant:
C
According Henry’slaw ofsolubilityofgases -
P=KHX
Withincreasein temperatureincreases,thus ‘x’ decreases given thepartialpressureofO2remainssame.
Question 3. In an experiment on depression of freezing point, two beakers leveled as A and B were filled with 1000 g. of water each. 40 g and 80 g of a certain non-electrolytic solute were added to both the beakers and freezing point of the solutions were noted as TfA and TfB respectively. Now both the solutions were poured into an empty beaker C and the freezing point of the mixed solution is found to be TfC . What is the ratio TfA : TfB : TfC?
Explanation:-
Answer: Option B. -> 2 : 4 : 3:
B
△Tf=Kfm
⇒0−Tf=Kfm
⇒=−Kfm
If 40 g of solute contains 'x'~moles of it then 80 g of the solut ill contains '2x' molesofit
mA=xWt.ofsolventing1000=x10001000=x
mB=2xWt.ofsolventing1000=x10001000=2x
mC=3xWt.ofsolventing1000=3x10001000=1.5x
Required ration TfA:TfB:Tfc=1.2:1.5=2:4:3
Explanation:-
Answer: Option B. -> 0.1 M Na3 PO4:
B
Depression in vapor pressure is a colligative property. More the no. of particles present in the solution lower will be the vapor pressure. consists of maximum ions hence it show lowest vapour pressure.
MoleculeNaClNa3Po4BaCl2SucroseNo.ofionspermolecule1+1=23+1=41+2=31
Question 5. A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at 250C. Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mm Hg at 250C. Calculate the molecular weight of the solute.
Explanation:-
Answer: Option D. -> 0.478:
D
Mole fractionof pentane invapourphase = Y1=P1P1+P2=P01X1P01X1+P02X2
=440×15440×15+120×15=0.478
Explanation:-
Answer: Option B. -> Benzene-methanol:
B
(a) Water-Nitric acid solution shows negative deviation from Raoult's law due to hydrogen bonding
(b) Benzene-methonol solution shows positive deviation due to formation hydrogen bonding
(c) Water-HCl solution shows negative deviation due to formation of hydrogen bonding
(d) Acetone-Chloroform solutions shows negative deviation due to formation of hydrogen bonding
it is prudent to remember some of these examples of solutions showing positive and negative deviations from Rault's law along with examples of ideal solutions
NegativedeviationfromRault′slawPositivedeviationfromRault′slawidealsolutions∙Chloroform−Acetone∙Water−Nitricacid∙Water−HCl∙Phenol−Aniline∙Benzene−Methylalcohol∙carbondisulfie−Acetone∙Chloroform−ethanol∙Water−Ethylalcohol∙Benzene−Toluene∙n−hexane−n−heptane∙Ethylbromide−Ethyliodide∙ChloroBenzene−bromoBenzene
Explanation:-
Answer: Option B. -> Both A and R are true and R is not a correct explanation of A:
B
Molalitydoesnotchange with change in temperature, because it is defined as no.of.molesofsolutewt.ofsolventinkg . Neither no. of moles nor the wt. of solvent is changing with change in temperature. Hence (b) is the correct option.
Question 9. Ratio of Henry’s constant, KH of two inert gases A & B is 2 at a given temperature. A mixture containing 4 moles of gas A and 8 moles of gas B is bubbled through water covered from top and solution is allowed to attain equilibrium. Assuming the absence of any other gases, what will be the ratio of mole fractions of gases in the solution.
Explanation:-
Answer: Option B. -> 8:9:
B
Let ‘a’ and ‘b’ moles of gases dissolve in water while rest accumulates over the solution to create partial pressure.
From Henry’s Law –
p=KHX
⇒p1P2=KH1KH2×X1X2
⇒4−a8−b=12×ab
⇒4−a/b8−1=12×ab
⇒ab=4×29=89
Hence Answer is option (b)
Explanation:-
Answer: Option A. -> 0.806 M, 4.83 N, 0.825 m:
A
Molecular wt. of Al2(SO4)3=2×27+3×(32+4×16)=342g/mole
Equivalent wt. of Al2(SO4)3=Eq.wt.ofAl3++Eq.WtofSO2−4=273+962
=57geq
No. of. equivalent per mole =34257=6
Let volume of solutions = 1 L
Wt. of solutions = V×density=1000×1.253=1253g
Wt. of solute = 1253×22%=257.66
Moles. of solute = 275.66342=0.806
Wt. of solvent = 1253−257.66=977.34
Molarity = molesofsoluteVolumeofsolution=0.806M
Normality = 6×Molality=6×0.806=4.836N
Molalit = MolesofsoluteWt.ofsolventink.g=0.806977.34/1000=0.825m