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87151.

A solution of glycerol (C_(3) H_(8) O_(3)) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42^(@) C while pure water boils at 100^(@)C. What mass of glycerol was dissolved to make the solution ? (K_(b)"for water " = 0.512 K "kg mol"^(-1))

Answer»

SOLUTION : `(W_(B))` = Mass of GLYCEROL = ?
`(W_(A))` = Mass of solvent = 500 g
Solution was prepared by dissolving some glycerol in water.
(100.42-100)
`Delta T_(b)= 0.42^(@)`C
`K_(b) = 0.51 " kg mol"^(-1)`
`Delta T_(b) = K_(b) xx `Molality.
Molality (m) = `(W_(b) xx 1000)/(W_(A) xx M_(B))`
Molecular wt.of Glycerol `(M_(B)) C_(3) H_(8) O_(3)`
= ` (12 xx 3) + (1 xx 8) + (16 xx 3)`
36 + 8 + 48 = 92
`Delta T_(b) = (W_(B) xx 1000)/(500 xx 92 ) xx K_(b)`
`0.42 = (0.512 xx W_(B) xx 1000)/(500 xx 92 )`
`W_(B)= (0.42 xx 500 xx 92 )/(0.512 xx 1000)`
`W_(B) = (19320)/(512)`
`W_(B) `= 37.88 g
Weight of glycerol = 37.88 g
Discussion
87152.

A solution of glucose (Molar mass ="180 g mol"^(-1)) in water has a boiling of 100.20^(@)C. Calculate the freezing point of the same solution. Molal constants for water K_(f) and K_(b) are "1.86 K kg mol"^(-1) and 0.512 K kg "mol"^(-1) respectively.

Answer»


SOLUTION :`m=(DeltaT_(B))/(K_(b))=(0.20)/(0.512)=0.39""DeltaT_(F)=K_(f)m=1.86xx0.39=0.73^(@)`
Discussion
87153.

A solution of glucose (Molar mass = 180 g "mol"^(-1) ) in water has a boiling point of 100.20^@C. Calculate the freezing point of the same solution. Molal constants for water K_f and K_b are 1.86 K kg "mol"^(-1) and 0.512 K kg "mol"^(-1)respectively.

Answer»

Solution :Given that the boiling point `(T_b)` of GLUCOSE solution = `100.20^@C`
APPLY the RELATION :
` Delta T_b = K_b xx m`
`m = (Delta T_b)/(K_b) = (0.20)/(0.512)`
m = 0.390 mol/kg
Applying it to the freezing point
` Delta T_f = K_f xx m`
` Delta T_f = 1.86 K kg "mol"^(-1) xx 0.390 "mol" kg^(-1) = 0.725 K`
Freezing point of the solution = 273.15 K - 0.725 K = 272.425 K
Discussion
87154.

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution ? If the density of solution is 1.2 g mL^(-1), then what will be the molarity of the solution ?

Answer»

SOLUTION :0 % w / w solution of glucose in water means that 10 g glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 - 10) g = 90 g of water.
Molar mass of glucose `(C_(6)H_(12)O_(6))`
`= 6xx12+12xx1+6xx16`
`= 180 g MOL^(-1)`
Then, number of MOLES of glucose
`= (10)/(180)mol`
= 0.056 mol
Molarity of solution `= (0.056 mol)/(0.09 KG)=0.62 m`
Number of moles of water `= (90 g)/(18 g mol^(-1))=5 mol`
Mole fraction of glucose `(x_(g))=(0.056)/(0.056+5)=0.011`
And, mole fraction of water `X_(w)=1-x_(g)`
`=1-0.011`
= 0.989
If the ensity of the solution is 1.2 kg `mL^(-1)`, then the volume of the 100 g solution can be given as,
`= (100 g)/(1.2 g mol^(-1))`
= 83.33 mL
`= 83.33xx10^(-3)L`
`therefore` Molarity of the solution `= (0.056 mol)/(83.33xx10^(-3)L)`
= 0.67 M.
Discussion
87155.

A solution of glucose in water is labelled as 10% w/w, what would be the molality and molefraction of each component in the solution ? If the density of the solution is 1.2 g mL^(-1) , thenwhat shall be the molarity of the solution ?

Answer»

Solution :10% w/w means 10 g glucose is PRESENT in 100 g solution,
Mass of water = 100 – 10 = 90 g = 0.09 kg 10
10 g glucose = 10/180 mol = 0.0555 mol
Number of moles in 90 g`H_2O = 90/18 = 5` moles
Molality of solution =0.0555 mol / 0.090 kg= 0.617 m
Mole fraction of glucose `= (0.0555)/(5 + 0.0555) = 0.01`
mole of fraction of `H_2O = 1 - 0.01 = 0.99`
Volume of 100 g solution = 100/1.2 mL = 83.33 mL = 0.08333 L
` THEREFORE ` molarity `= (0.0555)/(0.08333L) = 0.67 M`
Discussion
87156.

A solution of glucose in water is labelled as 10% w/w. What would be the molality and mole fraction of each component in the solution? If the density of the solution is "1.2 g mL"^(-1). Then what shall be the molarity of the solution?

Answer»

Solution :10 g glucose is present in 100 g solution, i.e., 90 g of water = 0.090 kg of water
`"10 g of glucose "=(10)/(180)" mol = 0.0555 mol,90 g "H_(2)O=(90)/(18)="5 MOLES"`
`"MOLALITY"=("0.0555 mol")/("0.090 kg")=0.617 m"`
`"x (Glucose)"=(0.0555)/(5+0.555)=0.01, x(H_(2)O)=1-0.01=0.99`
`"100 g solution"=(100)/(1.2)mL = "83.33 mL = 0.08333 K ,Molarity "=("0.0555 mol")/("0.08333 L")="0.67 M".`
Discussion
87157.

A solution of glucose at 27^(@)C has an osmotic pressure equal to 3xx10^(3) Pa. If molecualr weight of glucose is 180, then the number of grams of glucose present in one litre of solution is

Answer»

0.216
1.52
0.324
`1.80`

ANSWER :A
Discussion
87158.

A solution of FeCl_(3) in water acts as acidic due to

Answer»

Hydrolysis of `FE^(3+)`
Acidic impurities
Dissociation
Ionisation

Solution :`FeCl_(3) + 3H_(2)O hArr Fe(OH)_(3) + 3HCl`. Strong ACID and weak BASE.
Discussion
87159.

A solution of Fe^(2+) is titrated potentiaometrically using Ce^(4+) solution . Calculate the EMF of the redox electrode thus formed when a. 50% of Fe^(2+) is titrated b. 90% of Fe^(2+) is titrated c. 110% titration is done Given :E^(c-)._(Fe^(2+)|Fe^(3+))=-0.77V and Fe^(2+)Ce^(4+)rarrFe^(3+)+Ce^(3+),K=10^(14)

Answer»

Solution :`a.` During titration, the REDOX electrode is `Pt|Fe^(2+),Fe^(3+)` electrode
`E=E^(c-)._(cell)-(0.059)/(n)log.([Fe^(3+)])/([Fe^(2+)])`
`Fe^(3+) +e^(-) RARR e^(2+)`
At `50%` titration, `[Fe^(3+)]=[Fe^(2+)]`
`:. E=E^(c-)=-0.77V`
`b.` At `90%` titration, `([Fe^(3+)])/([Fe^(2+)])=(90)/(10)`
`{:(,Fe^(2+),rarr,Fe^(3+),+e^(-)),(Initial,90,,10,),(Fi nal,10,,90,):}`
`:. E=-0.77-0.059log 9 =-0.826 V`
`c.`
At `100%` titration, the electrode becomes,
`Pt|Ce^(3+),Ce^(4+)` electrode,where`([Ce^(4+)])/([Ce^(3+)])=(10)/(100)`
Cell reaction `:`

At equilibrium, `E_(cell)=0`
`E=E^(c-)._(cell)-(0.059)/(1) log K `
`E^(c-)._(cell)=(0.059)/(1) log 10^(14)`
`=0.826`
`:. x-0.77=0.822impliesx=1.596`
`:. E^(c-)._(Ce^(4+)//Ce^(3+))=1.596V`
`:.E^(c-)._(Ce^(3+)//Ce^(4+))=-1.596V`
So at `110%` titration,
`E_(Ce^(3+)//Ce^(4+))=E^(c-)._(Ce^(3+)//Ce^(4+))-0.059log .([Ce^(4+)])/([Ce^(3+)])`
`=-1.596-0.0596 log ((10)/(100))`
`=-1.596+0.059=-1.537`
Discussion
87160.

A solution of ethanol in water is 10% by volume. If the solution and pure ethanol have densities of 0.9866g//"cc" and 0.785g//"cc" respectively, find the per cent by weight.

Answer»

Solution :VOLUME of ethanol `=10mL`, VOL of soultion `=100ML`
WEIGHT of ethanol `=` volume `xx` density
`=10xx0.785=7.85g`
Weight of solution `=100xx0.9866=98.66g`
`:.` weight per CENT `=(7.85)/(98.66)xx100=7.95%`
Discussion
87161.

A solution of electrolyte in water, when electrolysed , liberated H_(2) at the cathode and Cl_(2) at the anode. The electrolyte must be

Answer»

`H_(2)SO_(4(AQ))`
`CuCl_(2(aq))`
`NaCl_((Q))`
`Na_(2)SO_(4(aq))`

Answer :C
Discussion
87162.

A solution of CuSO_(4) is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? (Molar mass of Cu=63.5 g mol^(-1))

Answer»

Solution :Current=1.5 amperes, time=10min=`10xx60s=600s`
`THEREFORE`Quantity of ELECTRICITY passed=Current (amp)`XX`time (s)`=1.5xx600`coulombs=900C
The reaction occuring at the cathode is: `Cu^(2+)+2E^(-)toCu`
Thus, 2F, i.e., `2xx96500C` deposit Cu=10 mol=63.5 g
900C will deposit `Cu=(63.5)/(2xx96500)xx900=0.296g`
Discussion
87163.

A solution of d-glucose in water rotates the plane polarised light :

Answer»

To the right
to the LEFT
To either SIDE
None

ANSWER :A
Discussion
87164.

A solution of D (+) - 2-chloro-2-phenylethane in toluene racemises slowly in the presence of small amount of SbCl_(5), Due to the formation of

Answer»

CARBANION
Carbene
Free radical
Carbocation

SOLUTION :
Discussion
87165.

A solution of CuSO_(4) is electroysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ? (Molar mass of Cu=63.5 g//mol)

Answer»

Solution :`I=1.5" AMPERE"`
`"Time "=10xx60s=600s`
`Q=Ixxt`
`=1.5xx600=900C`
`Cu^(2+)+2e^(-)rarrCu(s)`
2F amount of electricity DEPOSIT copper = 63.5 G
`"900 C amount of electricity deposit copper"=(63.5xx900)/(2xx96500)`
`=0.296g`
Discussion
87166.

A solution of CuSO_(4) is elecrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode ?

Answer»

Solution :Quantity of ELECTRICITY Q=It
`THEREFORE Q=10xx60sxx1.5A|"where, "I=1.5A`
`t=10xx60s`
Deposition reaction of copper and its mole
Cathode reaction: `Cu_((AQ))^(2+)+2e^(-)toCu_((s))`
According to stoichiometry, for production of 1 mol of Cu, 2 mol of electron=2F electricity is USED.
So, 2F=`2xx96500F` will gives 1 mol
=63.5 g Cu
So, 900 C will REDUCED Cu`=(900Cxx63g)/((2xx96500)C)`
`=0.2938g" "Cu`
Discussion
87167.

A solution of CuCl inNH_4OH is used to measure title amount of which gas is a sample by simply measuring change in volume:

Answer»

`CO_2`
`H_2`
CO
All

Answer :C
Discussion
87168.

A solution of CuSO_4 in water will :

Answer»

Turn red litmus BLUE
Turns blue litmus red
Show no EFFECT on litmus
Decolourise litmus

Answer :B
Discussion
87169.

A solution of crab hemocyanin, a pigmented protein extracted from crabs, wasprepared by dissolving 0.75 g in 125 cc of an aqueous medium. At 4^@C , an osmotic pressure rise of 2.6 mm of the solution was observed. The solution had a density of 1 g/cc. Determine the molecular weight of the protein. Solve this problem in cgs units.

Answer»

SOLUTION :`5.4 XX 10^5 ` g/mole
Discussion
87170.

A solution of Cr(NO_3)_3 slowly turns green when concentrated HCl is added to it . It is due to the formation of:

Answer»

`CrCl_3`
`Cr_2O_3`
`CrO`
CHLORO COMPLEX `[CR(H_2O)_6]Cl_3`

ANSWER :D
Discussion
87171.

A solution of copper sulphate is electrolysed using a current strength of 3 amp to deposite 60 grams of copper. What is the time taken for the electrolysis?

Answer»

Solution :Deposition ofcopper is given as :
`Cu_((aq))^(2+)+2e^(-)rarrCu_((s))`
2 Faradays or 193000 coulomb can DEPOSIT 63.5 grams of copper. QUANTITY of ELECTRICITY (Q) required to deposit 60 grams of copper `=(60)/(63.5)xx19300 = 182362` coulomb.
Time taken for the electrolysis `(t)=(Q)/(i)=(182362)/(3)=60787` sec `=16.9` hrs.
Discussion
87172.

A solutionof copper sulphate may be kept safely in the container made up of

Answer»

Fe
AG
Zn
AI

Solution :Since Ag is less reactive than Cu THEREFORE it does not DISPLACE Cu from `CuSO_(4)` while other METALS are more reactive, they displace Cu from `CuSO_(4).`
Discussion
87173.

A solution of [Co(NH_(3))_(4)I_(4)]Cl when treated with AgNO_(3) gives a white precipitate.What should be the formula of isomer of the dissolved complex that gives yellow precipitate with AgNO_(3). What are the above isomers called ?

Answer»

Solution :`(i)` A solution of `[Co(NH_(3))_(4)I_(2)]Cl` when treated with `AgNO_(3)` gives a WHITE precipitate, because `Cl^(-)` ion is counter ion.
`(ii)` FORMULA of isomer of the dissolved complex that gives yellow precipitate with `AgNO_(3)` is, `[Co(NH_(3))_(4)Cl I]I` BECUASE `I^(OPLUS)` is counter ion
`(iii) [Co(NH_(3))_(4)I_(2)]Cl` and `[Co(NH_(3))_(4)Cl I ] I` both are ionisation isomers.
Discussion
87174.

A solution of concentration C gequiv/L, has a specific resistance R. The equivalent conductance of the solution is

Answer»

R/C
C/R
`(1000)/(RC)`
`(1000R)/(C)`
Discussion
87175.

A solution of colourlesssalt H on holing with excess NaOH produces a non-flammable gas .The gas evolution ceses after sometime. Upon addition of Zn dust to the same solution the gas evolutionrestarts .The colourless salt(s) H is (are)

Answer»

`NH_(4)NO_(3)`
`NH_(4)NO_(2)`
`NH_(4)CI`
`(NH_(4))_(2)SO_(4)`

Solution :`NH_(4)ON_(3) + NaOH rarr NH_(3) + NaNO_(3) + H_(2)O`
`7NaOH + NANO + 4Zn rarr 4Na_(2)ZnO_(2) + NHO_(3) + 2H_(2)O`
`NH_(4)ON_(2) + NaOH rarr NaNO_(2) + NH_(3) + H_(2)O`
`3Zn + 5NaOH + NaNO_(2)+ rarr 3Na_(2)ZnO_(2) + NH_(3)O`
Discussion
87176.

A solution of colourless salt H on boiling with excess NaOH produces a nonflammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are)

Answer»

`NH_(4)NO_(3)`
`NH_(4)NO_(2)`
`NH_(4)Cl`
`(NH_(4))_(2)SO_(4)`

Solution :`NH_(4)NO_(3)+NaOHtoNH_(3)+NaNO_(3)+H_(2)O`
`7NaOH+NaNO_(3)+4Znto4Na_(2)ZnO_(2)+NH_(3)+2H_(2)O`
`NH_(4)NO_(2)+NaOHtoNaNO_(2)+NH_(3)+H_(2)O`
`3Zn+5NaOH+NaNO_(2)to3Na_(2)ZnO_(2)+NH_(3)+H_(2)O`
HENCE,(A) and (B) are the correct answers.
Discussion
87177.

A solution of colourless salt H on boiling with excess NaOH produces a nonflammable gas.The gas evolution ceases after some time.Upon addition of Zn dust to the same solution, the gas evolution restarts.The colourless salt(s) H is (are) :

Answer»

`NH_(4)NO_(3)`
`NH_(4)NO_(2)`
`NH_(4)Cl`
`(NH_(4))_(2)SO_(4)`

Solution :All ammonium salts on reaction with alkali produce ammonia.The nitrate and nitrite also on reduction with nascent hydrogen (produced by the reaction of zinc and sodium hydroxide) produce the ammonia gas according to the following REACTIONS.
`NH_(4)NO_(2)+NaOH to NaNO_(2)+NH_(3)+H_(2)O`
`NH_(4)NO_(2)+NaOH to NaNO_(3)+NH_(3)+H_(2)O`
`NaNO_(2)+6[H]overset(Zn//NaOH) to NaOH+NH_(3)+H_(2)O`
`NaNO_(3)+8[H]overset(Zn//NaOH) to NaOH+NH_(3)+2H_(2)O`
So options `(A)` and `(B)` are CORRECT.
Discussion
87178.

A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are)

Answer»

`NH_(4)NO_(3)`
`NH_(4)NO_(2)`
`NH_(4)CL`
`(NH_(4))_(2)SO_(4)`

Solution :The colourless salt H may be either `NH_(4)NO_(3) or NH_(4)NO_(2)` and the non-inflammable gas is `NH_(3)`
(a) `{:(NH_(4)NO_(3) + NaOH overset(Delta)rarr underset("Non-inflammable gas")(NH_(3)) + NaNO_(3) + H_(2)O),(4 Zn + 7 NaOH + NaNO_(3) overset(Delta)rarr underset("SOD. ZINCATE")(4Na_(2)ZnO_(2)) + underset("Gas evolution restarts")(NH_(3)) + 2H_(2)O):}`
(b) `{:(NH_(4)NO_(2) + NaOH overset(Delta)rarr NH_(3) + NaNO_(2) + H_(2)O),(3 Zn + 5 NaOH + NaNO_(2) overset(Delta)rarr 3Na_(2)ZnO_(2) + NH_(3) + H_(2)O):}`
Discussion
87179.

A solutionof coloured salt if bollingwithexcess NaOHproducesa non flammable gas .The gas evolvurationcoases aftersometime .Upon additionof Zn dust to the samesolution , the gas evolationrestart .The colourless salt (s) H is/are

Answer»

`NH_(4)NO_(3)`
`NH_(4)NO_(2)`
`NH_(4)CI`
`(NH_(4))_(2)SO_(4)`

Solution :The gas is ammonimia and the saltcan be nitrate or nitrit as SUGGESTED by the following reaction
`NH_(4)NO_(3) + NAOH rarr NaNO_(3) + NH_(3) + H_(2)O`
`NaNO_(3) + 8[H] rarr NaOH + NH_(3) + 2H_(2)O`
`NH_(4)NO_(2) + NaOH rarr NaNO_(3) + NH_(3) + H_(2)O`
`NaNO_(2) + 6[H] rarr NaOH + NH_(3) + H_(2)O`
Discussion
87180.

A solution of Benzylamine, Cyclohexanol and Picric acid in ethyl acetate was extracted initially with a saturated solution of NaHCO_(3)to give fraction A. The left over organic phase was extracted with chloroform with KOH give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C contain respectively :

Answer»

PICRIC ACID, BENZYLAMINE, CYCLOHEXANOL
Benzylamine, Picric acid, Cyclohexanol
Picric acid, Cyclohexanol, Benzylamine
Benzylamine, Cyclohexanol, Picric acid

Solution :
Discussion
87181.

A solutionofan organic compundis prepared by dissolving34.2 g in 500 g ofwater.Calculate the molecularmass of the compoundandfreezingpoint of thesolution . Given that k_(b) for water= 0.52 K m^(-1) , b.pt of solution= 100.14^(@)C , K_(f) for water= 1.87 Km^(-1) .

Answer»


Solution :`Delta T_(B) = 100.104 - 100 = 0.104^(@)`
`M_(B) = (0.52 xx 34.2 xx 1000)/(0.14 xx 500)= 342`
`Delta T_(f) = (1.87 xx 34.2 xx 293)/(342 xx 500) = 0.374`
`therefore ` FREEZING POINT`= - 0.374^(@)C`
Discussion
87182.

A solution of cane sugar at 27^@Cdevelops an osmotic pressure of 4.93 atm. Calculate the f.p. of this solution (molecular depression constant for 100 g of water is 18.6).

Answer»

SOLUTION :`0.372^@C`
Discussion
87183.

A solution of chlorine in water contains:

Answer»

HOCL only
HCl only
HCl and HOCl
HCl and HOCl and chlorine

Answer :D
Discussion
87184.

A solution of CaCl_(2) is mol/litre, then the moles of chloride ions in 500 mL will be

Answer»

0.25
0.5
0.75
1

Solution :`M=n/(V(ML)) xx 1000`
`0.5=n/500 xx 1000`
`n=0.5/2=0.25` moles of `CaCI_(2)`
`As CaCI_(2) hArr Ca^(2+) + 2 CI^(-)`
Discussion
87185.

A solutionof an organiccompound was prepared bydissolving6.8 gin 100 g of water. Calculatethe osmoticpressureof thissolution at 298 Kwhenboilingpointof solutionis 100.11^(@)C . GivenK_(b) for water= 0.52 Km^(-1) and R = 0.082 litre atm k^(-1) mol^(-1)

Answer»

Solution :Molecularmass can be CALCULATED from boilingpointdata as :
`M_(B) = (k_(b) xx w_(B) xx 1000)/(Delta T_(b) xx w_(A))`
`w_(B) = 6.8 G , w_(A) = 100 g, k_(b) = 100.11 - 1000 = 0.11`
`k_(b) = 0.52 K m^(-1)`
`M_(B) = (0.52 xx 6.8 xx 1000)/(0.11 xx 100)`
` = 321.45`
Now `"" pi = (nRT)/( V) , n = (6.8)/(321.45)`
`V = 100 g ~~ 100nl = 0.1 L`
`R = 0.082 L " atm " mol^(-1) K^(-1)`
`R= 0.082 L " atm " mol^(-1) k^(-1)`
`T = 298 K`
` pi = (6.8 xx 0.082 xx 298)/(321.45 xx 0.1)`
`=5.17` atm.
Discussion
87186.

A solutionof anorganiccompoundis preaped by dissolving68.4 g in 1000 g ofwater .Calculatepressureof thesolutionat 293 K when elevationin boiling point is 0.104 andK_(b)for wateris 0.53 Km^(-1) .

Answer»


Solution :`M_(B)= (0.52 xx 68.4 xx 1000)/(0.104 xx 500) = 342`
` pi= (68.4 xx 0.082 xx 293)/(342 xx 500) = (V = 1000 g=1 L)`
` = 4.80` atm.
Discussion
87187.

A solution of a weak acid (Ka = 10^(-5) ) has a molarity of (M/5). lOmL of this solution is neutralised completely with a NaOH solution of molarity (M/20). At the neutralisation point, concentration of H_3O^+ ions (mol*L^(- 1)) is-

Answer»

`4.39xx10^(-5)`
`1.25xx10^(-6)`
`7.02xx10^(-6)`
`1.58xx10^(-9)`

ANSWER :D
Discussion
87188.

A solution of an organic compound is prepared by dissolving 68.4 g in 1000 gof water. Calculate the molecular mass of the compound and osmotic pressure of the solution at 293 K when elevation of b.pt is 0.104 and K_(b) for water is "0.52 K mol"^(-1).

Answer»


SOLUTION :`DeltaT_(b)=K_(b)xxm THEREFORE m=DeltaT_(b)//K_(b)=0.104//0.52 = "0.2 mol Kg"^(-1)`
`"THUS, 0.2 moles = 68.4 g"therefore"1 mole "=(68.4)/(0.2)=342," i.e,Molar MASS = 342 g mol"^(-1)`
`"Mass of solution = 1068.4g. Taking it as a s dilute solution, its density "~="1 g CM"^(-3)." Hence, Volume "=1068.4cm^(3)=1.0684L`
`therefore"Molar concentration "=("0.2 mol")/("1.0684L")="0.187 mol L"^(-1)`
`pi=CRT-0.187"mol L"^(-1)xx0.082"L atm K"^(-1)"mol"^(-1)xx"293 K = 4.50 atm."`
Discussion
87189.

A solution of an organic compound is prepared by dissolving 34.2 g in 500 g of water. Calculate the molar mass of the compound and freezing point of the solution. Given that K_(b) for water ="0.52 K mol"^(-1) B.pt of solution =100.104^(@)C.K_(f) for water =1.87"K mol"^(-1).

Answer»


Solution :`DeltaT_(b)=0.104^(@),w_(2)=34.2g, w_(1)=500 g, K_(b)=0.52Kmol"^(-1)`
`therefore""M_(2)=(1000K_(b)w_(2))/(w_(1)xxDeltaT_(b))=(1000xx0.52xx34.2)/(500xx0.104)="342 g mol"^(-1)`
`DeltaT_(F)=K_(f)xxm=(1.87)XX((34.2)/(342)xx(1)/(500)xx1000)=0.374""therefore"F.pt "=-0.374^(@)C`.
Discussion
87190.

A solution of a substances is titrated against astrong base ( or acid ) ,volume V of strong base (or acid) is plotted against pH of the solution (as shown in figure) .The substances could be :

Answer»

`Na_(2)CO_(3)`
Ethylene diamine
`H_(2)C_(2)O_(2)`
`CH_(2)(COOH)_(2)`

Solution :Given graph is PLOTTED for the titration of weak diprotic ACID with base.
Discussion
87191.

A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783 g. Calculate the equivalent mass of the metal.

Answer»

Solution :Given `I = 0.15` amperes
`t = 150` mins `implies t = 150 XX 60 "sec" implies t = 9000` sec
`Q = I t implies Q = 0.15 xx 9000` coulombs `implies Q = 1350` coulombs
Hence, `135` coulombs of electricity DEPOSITE is EQUAL to `0.783 g` of metal.
`:. 96500` coulombs of electricity , `(0.783 xx 96500)/(1350) = 55.97` gm of metal
Hence equivalent mass of the metal is `55.97`.
Discussion
87192.

A solution of a salt of metal was electrolysed for 150 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783g. Calculate the equivalent mass of the metal.

Answer»

Solution :Given, `I=0.15` amperes
`t=150" MINS" rArr t=150xx60 SEC rArr t=9000 sec`
`Q=It rArr Q=0.15xx9000" coulombs" rArr Q=1350` coulombs
Hence, 135 coulombs of electricity deposit is EQUAL to 0.783g of METAL.
`therefore` 96500 coulombs of electricity, `(0.783xx96500)/(1350)=55.97g` of metal
Hence equivalent MASS of the metal is 55.97
Discussion
87193.

A solution of a salt of metal was electrolysed for 15 minutes with a current of 0.15 amperes. The mass of the metal deposited at the cathode is 0.783 g. Calculate the equivalent mass of the metal.

Answer»

Solution :`I=0.15` amperes
times `t=15 times 60s=900s`
mass `m=0.783g`
`m=2It`
`z=m/(It)`
`Q=I times t=0.15 times 900 C`
Amount of the metal DEPOSITED by 1 C `=0.783/135=0.0058`
Electrochemical equivalent `Z=5.8 times 10^(-3)GC^(-1)`.
Discussion
87194.

A solution of a salt of a metal eletrolysed for 150 minutes with a current of 0.15 gm. The equivalent weight of the metal is

Answer»

55.97 gm
65.97 gm
75.97gm
85.97 gm

SOLUTION :Time(t)=150min=9000sec
Current(I)=0.15A
WEIGHT of metal (w)=0.783g.
We know `Q=Ixxt=0.15xx9000=1350C.` SINCE 1350 C
of electricity will DEPOSITED`(0.783xx96500)/(1350)=55.97g`.
Discussion
87195.

A solution of a salt of a metal was electrolysed for 150 minutes by passing 0.15 A current. The weight of the metal deposited was 0.783 g. The specific heat of the metal is 0.057" cal"//gK. The atomic mass X of the metal is :

Answer»

`111.80" G"//mol`
`52.2" g"//mol`
`200" g"//mol`
`250" g"//mol`

Solution :(a) According to Dulong & Petits Law,
Approximate Atomic mass`=(6.4" Cal"//K)/(0.057"Cal"//gK)=112.28" g"`
Quantity of charge passed
`=(0.5" AMP")xx(150xx60 s)`
`=1350" amp"-s=1350" C"`
1350 C charge deposit metal =0.738 g
96500 C charge deposit metal
`=((0.738g))/((1350c))xx(96500c)=55.9 g`
`:.` Equivalent mass of metal=55.97 g
Approximate valency`=("App.Atomic mass")/("Equivalent mass")`
`=((112.28g))/((55.97 g))=2.006`
SINCE valency is a whole number,
Exact valency =2
Exact atomic mass`="Eq.mass"xx"valency"`
`55.97xx2=111.94 g`
Discussion
87196.

A solution of a salt in water, on addition of dilute HCI, gives a white precipitate, which is soluble in hot water. The salt contains

Answer»

 `Fe^(2+)`
`HG^(2+)`
`AG^(+)`
 `PB^(2+)`

ANSWER :D
Discussion
87197.

A Solution of a non-volatile solute in water freezes at -0.30^@C. The vapour-pressure of pure water at 298K is 23.51 mm Hg and K_f for water is 1.86 K. kg "mol"^(-1) Calculate the vapour-pressure of this solution at 298K.

Answer»

23.4 mm
24.8 mm
34.8 mm
40 mm

Answer :A
Discussion
87198.

A solution of a non-volatile solute in water freezes at -0.30^(@)C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and K_(f) for water is 1.86 degree//molal. Calculate the vapourpressure of this solution at 298 K.

Answer»

SOLUTION :23.44 MM
Discussion
87199.

A solution of a non-volatile solute in ethanol (B.P of ethanol=78.4^(@)C) has a vapour pressure of730mm of Hg at 78.4^(@)C . To what temperature, the above solution must be heated have the pressure of 760mm of Hg? K_(b) of ethanol =1.22k Kg "mol"^(-1). Express temperature of 10x. x is ____

Answer»


ANSWER :8
Discussion
87200.

A solution of a metal ionwhen tretedwith KI gives a redprecipitatewhichdissolves in excess KI to give a colourless solution Moreever the solutionof metal ion ontreatmentwith a solution of cobalt (II) thoiocyamategivesriseto a deep blue crystallineprecipitate.THe metal ionis

Answer»

`Pb^(2+)`
`HG^(2+)`
`CU^(2+)`
`Co^(2+)`

Solution :`Hg^(2+) + 2I^(Theta) RARR HgI_(2) darr `(Scarlet redprecipitate )
`HgI_(2) + 2I^(theta) rarr [HgI_(4)]^(2+)`
`Hg^(2+) + Co(SCN)_(2) rarr underset("DEEP blue crystalaline")(Hg(SCN)_(2)) + Co^(2+)`
Discussion