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91851.

7 . 46 g of KCl washeated with excess of MnO_(2) and H_(2)SO_(4) the gas so produced was thenpassed througha solution of Kl. Calculatethe weight of iodine .

Answer»


Solution :`HCI underset(H_(2)SO_(4))overset(MnO_(2))toCl_(2)overset(KI)to I_(2): ` (MOLES of `Cl_(2)`= moles of `I_(2)` )
Discussion
91852.

A 0.2 molal aqueous solution of a weak acid (HX) is 20 per cent ionised. The freezing point of this solution is: (K_f = 1.86 K/m for water)

Answer»

`-0.45^@C`
`-0.90^@C`
`-0.31^@C`
`-0.53^@C`

ANSWER :A
Discussion
91853.

6mole of an ideal gas expand isothermally and reversibly from a volume of 1 litre to a volume of 10 litre at 27° C .The maximum work is done

Answer»

47kJ
100kJ
0
34.46kJ

Answer :D
Discussion
91854.

A 0.2 molal aqueous solution of a weak acid (HX) is 20 percent ionised. The freezing point of this solution is: (K_f = 1.86 K/m for water)

Answer»

`-0.45^@C`
`-0.90^@C`
`-0.31^@C`
`-0.53^@C`

ANSWER :A
Discussion
91855.

6I^(-)(aq)+BrO_3^(-)(aq)+6H^(+)(aq)to3I_3(aq)+Br^(-)(aq)+3H_2O(l) These data were obtained when this reaction was studied. {:([T]","M,[BrO_3^(-)]","M,[H^+]","M,"Reaction rate mol"L^(-1)s^(-1)),(0.0010,0.0020,0.010,8.0xx10^(-5)),(0.0020,0.0020,0.010,1.6xx10^(-4)),(0.0020,0.0040,0.010,1.6xx10^(-4)),(0.0010,0.0040,0.020,1.6xx10^(-4)):} What are the units of the rate constant for this reaction ? 6I^(-)(aq)+BrO_3^(-)(aq)+6H^(+)(aq)to3I_3(aq)+Br^(-)(aq)+3H_2O(l)

Answer»

`s^(-1)`
`mol L^(-1) s^(-1)`
`L mol^(-1) S^(-1)`
`L^2 mol^(-2) s^(-1)`

SOLUTION :ORDER of reaction is 2 therefore unit FO K is `"mole"^1 Ls^(-1)`
Discussion
91856.

A 0.1M solution of sodium acetate was prepared. The K_(h)=5.6xx10^(-10) . Then :

Answer»

The degree of HYDROLYSIS is `7.48xx10^(-6)`
the `[OH^(-)]` concentration is `7.48xx10^(-3)M`
the `[OH^(-)]` concentration is `7.48xx10^(-6)M`
the pH is APPROXIMATELY 8.88

Answer :c,d
Discussion
91857.

6gm of silver salt of tribasic acid gives 4.32 gm silver on strong heating. The molar mass of acid is:

Answer»

126gm
129gm
123gm
252gm

Solution :`H_(3)Aoverset(AgNO_(3))rarrAg_(3)Aoverset(Delta)rarrAgdownarrow`
`POAC "on" Ag: 3xxn_(Ag_(3)A)=n_(Ag)`
`3XX(6)/(M.Wt_("Salt"))=(4.32)/(108)`
`M.Wt_("salt")=(3xx6xx108)/(4.32)=450`
M.Wt. ACID `=450-108xx3+3xx1`
M.Wt. acid=`450-107xx3=129`
Discussion
91858.

6g weak acid HA(molecular Wt. = 60 g/mol) is dissolved in water and formed 10 L solution, if Ka of HA is 10^(–8), then p^(OH) of solution is :-

Answer»

`5.0`
`8.0`
`9.0`
`10`

ANSWER :A
Discussion
91859.

A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g "mol"^(-1) ) has a freezing pointof 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g "mol"^(-1) ) per 100 g of solution ?

Answer»

SOLUTION :The following DATA is provided :
`Delta T_f = T_f^0 = 273.15 - 271 = 2.15 K`
Using the following relation and substituting the values, we GET
`Delta T_f = K_f XX m`
`2.15 = K_f xx 0.1539`
`K_f = (2.15)/(0.1539)`
` Delta T_f = K_f xx m = (2.15)/(0.1539)xx 5/180 xx 1000/95 = (10750)/(2631.69) = 4.08 K`
Freezing point = 273.15 - 4.08 = 269.07 K.
Discussion
91860.

6CO_(2)+6H_(2)O overset("Sunlight")rarr C_(6)H_(12)+6O_(2) Name the above process.

Answer»

SOLUTION :PHOTOSYNTHESIS.
Discussion
91861.

A 0.1539 molal aqueous solution of cane sugar (mol. Mass = "342 g mol"^(-1)) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. Mass = "180 g mol"^(-1)) per 100 g of solution.

Answer»


SOLUTION :`DeltaT_(F)=K_(f)m therefore K_(f)=(DeltaT_(f))/(m)=(273.15-271)/(0.1539)="13.97K/m, "DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2))=(1000xx13.97xx5)/(100xx180)=3.88`
`therefore"FREEZING point of the solution "=273.15-3.88K=269.27K.`
Discussion
91862.

._(6)C^(14) is formed from ._(7)N^(14) in the upper atmosphere by the action of the fundamental particle

Answer»

Positron
Neutron
Electron
Proton

Solution :The radioactive ISOTOPE `._(6)C^(14)` is produced in the atmosphere by the ACTION of cosmic ray NEUTRONS on `._(7)N^(14)`
`._(7)N^(14) + ._(0)n^(1) RARR ._(6)C^(14) + ._(1)H^(1)`
Discussion
91863.

A 0.100M aqueous solution of H_(2)SeO_(3) is titrated with 1.000M NaOH solution. At the point marked with a circle on the titration curve, which species represent at least 10% of the total selenium in solution ?

Answer»

`H_(2)SeO_(3)` only
Both `H_(2)SeO_(3)` and `HSeO_(3)^(-)`
`HseO_(3)^(-)` only
Both `HSeO_(3)^(-)` and `SeO_(3)^(2-)`

ANSWER :C
Discussion
91864.

A 0.1 N solution of Na_(2)CO_(3) is titrated with 0.1 N HCl solution. The best indicator to be used is

Answer»

POTASSIUM ferricyanide
Phenolphthalein
Methyl red
Litmus Paper

SOLUTION :As the RESULTING solution is ACIDIC.
Discussion
91865.

._(6)C^(12) and ._(1)T^(3) are formed in nature due to the nuclear reaction of neutron with

Answer»

`._(7)N^(14)`
`._(6)C^(13)`
`._(2)He^(4)`
`._(3)Li^(6)`

Solution :The BALANCED equation is
`._(7)N^(14) + ._(0)n^(1) RARR ._(6)C^(12) + ._(1)T^(3)`
`._(1)T^(3)` (Tritium) is a radioactive isotope of hydrogen
Discussion
91866.

A 0.1 N solution of an acid at room temperature has a degree of ionisation 0.1. The concentration of OH^(-) would be

Answer»

`10^(-12) M`
`10^(-11) M`
`10^(-9) M`
`10^(-2) M`

ANSWER :A
Discussion
91867.

6.90M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = 56 g mol^(-1))

Answer»

Solution :Mass of KOH = 30 g
`M = (n_(B))/(V(ml)) xx 1000`
`= (W_(B))/(M_(B) xx V(ml)) xx 1000 = (30)/(56 xx V) xx 1000`
`6.90 = (30 xx 1000)/(56 xx V)`
`V = (30 xx 1000)/(56 xx 6.90) = 81.43` mL
`D = (M)/(V)`
` = (100)/(81.43) = 1.28 g mL^(-1)`
Discussion
91868.

A 0.1-mole sample of NO_2 was placed in a 10-litre container and heated to 750 K. The total pressure of the equilibrium mixture as a result of the decomposition 2NO_2(g)to2NO(g) + O_2(g)was 0.827 bar. What is the value of K_pat this temperature? What amount of NO_2must be placed in this container to obtain an equilibrium concentration of NO_2of 0.1 mole per litre?

Answer»

SOLUTION :0.704, 1.61 MOLE
Discussion
91869.

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If K_(f) for water is 1.86^(@)C//m, the freezing point of the solution will be

Answer»

`-0.18^(@)C`
`-0.54^(@)C`
`-0.36^(@)C`
`-0.24^(@)C`

Solution :`{:(HA,RARR,H^(+),+,A^(-)),(1-ALPHA,,alpha,,alpha),(1-0.3,,0.3,,0.3):}`
`THEREFORE i=(1-0.3)+0.3+0.3=1.3`
`DeltaT_(f)=iK_(f)m=1.3xx1.86xx0.1=0.2418`
`T_(f)=0-0.2418^(@)C=-0.2418^(@)C~=-0.24^(@)C`.
Discussion
91870.

68.4 g of sugar (molecular weight = 342) is dissolved in 1000 g of water. What is (a) freezing point (b) boiling point (c) vapour pressure at 20^(@)C (d) osmotic pressure of the solution at 20^(@)C? The density of the solution at 20^(@)C is "1.024 g cm"^(-3). The vapour pressure of water at 20^(@)C is 17.633 mm. The K_(f) and K_(b) values for water are 1.873^(@) and 0.516^(@) respectively.

Answer»


Solution :Apply `Delta T_(f)=K_(f)m and DeltaT_(b)=K_(b)m,` Mass of solution = `1000+68.4=1068.4g`.
`"VOLUME "=(1068.4)/(1.024)CM^(3)`.
Discussion
91871.

A 0.1 Msolution of a weak acid HA is titrated by NaOH slowly. The titration curve observed is given. ka = 4 xx 10^(-5) at 25^(@)C. The correct observation for the curve will be:

Answer»

SOLUTION will MAKE buffer before equivalence point
`PH` of solution at equivalence point is `5.6`
`pH` of solution at equivalence point is 7
Salt hydrolysis cannot SUITABLE to predict the `pH` of equivalence point

Solution :Solution will make buffer before equation point since before equation point both `HA` amd `NaA` are present.
Discussion
91872.

68% aqueous nitric acid cannot be concentrated by further fractional distillation. Give reason.

Answer»

Solution :`68%` AQUEOUS nitric acid cannot be concentrated by further fractional distillation ALSO because it is an AZEOTROPIC in nature.
Discussion
91873.

A 0.1 M solution of an acid (density = 1.01 g/cc) is 4.5% ionised. Calculate the f.p. of the solution. The molecular weight of the acid is 300. K_f= 1.86.

Answer»

SOLUTION :`(-0.199^@C)`
Discussion
91874.

_(67)^(165)Ho is stable isotope. _(67)^(150)Ho is expected to distegrate by:

Answer»

`ALPHA`-EMISSION
`BETA`-emission
Positron emission
`GAMMA` -emission

Answer :C
Discussion
91875.

A 0.1 M solution of a weak acid HA is 1% dissociated. The approximate value of dissociation constant is :

Answer»

`1.0xx10^(-4)mol L^(-1)`
`0.1XX10^(-2)mol L^(-1)`
`1xx10^(-3)mol L^(-1)`
`0.1xx10^(-4)mol L^(-1)`

Solution :`K_(a)=c alpha^(2)=0.1xx(0.01)^(2)`
`=0.1xx10^(-4)`
Discussion
91876.

.^(64)Cu(T_(50)=12.8h) decays beta^(-) emission (38%), beta^(+) emission (19%) and electron capture (43%). Write the decay products and calculate partial half-lives for each of the decay processes.

Answer»


Answer :`t_(1)=33.68h`
`t_(2)=67.36h`
`t_(3)=29.76h`
Where `t_(1)`, `t_(2)` and `t_(3)` are the parallel half lives for `BETA^(-)` emission , `beta^(+)` emission and electron CAPTURE processes, respectively.
Discussion
91877.

A 0.1 M basic solution is required from Ca(OH)_(2) is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ions, Na^(+) and carbonate ions, CO_(3)^(2-) are respectively (Molar mass of Na_(2)CO_(3)=106"g mol"^(-1))

Answer»

`0.125M`
`0.25M`
`0.4M`
`0.5M`

Solution :As 1 mole of `CA(OH)_(2)` PRODUCES 2 moles of `OH^(-)` ions, therefore, if it were completely ionized.to get 0.1 M `OH^(-)` ions, `Ca(OH)_(2)` solution required WOULD have been = 0.05 M.
Discussion
91878.

65,500 columb = _______

Answer»

`6.28xx10^(18)` electrons
`6.023 XX 10^(23)` electrons
`6.023 xx 10^(23)` electrons
`6.28 xx 10^(27)` electrons

Solution :1 MOLE of electrons `=6.022xx10^(23) E^(-) =96500 C=1F`
Discussion
91879.

A 0.1 M aqueous of Na_(2)SO_(4) is diluted by adding water. Water will happen to the values of its conductance (G), conductivity (kappa), molar conductivity (wedge_(m)) and equivalent conductivity (wedge_(eq))?

Answer»

Solution :`kappa` decreases WHEREAS `G,wedge_(m) and wedge_(eq)` INCREASE with DILUTION.
Discussion
91880.

6.539 xx 10^(-2)of metallic Zn (65.39 amu) was added to 100 mL of saturated solution of AgCl. Calculate log ([Zn^(2+)] //[Ag^+]^2)Given Ag^(+) + e = Ag , E^@ = 0.80 VZn^(2+) + 2e = Zn, E^@ = 0.76VAlso find how many moles of Ag will be formed.

Answer»

SOLUTION :`52.79 , 10^(-6)` MOL
Discussion
91881.

""^(64)Cu(T""_(50)=12.8 year) decays beta-emission (38%), beta""^(+) decayproducts and calculate partial half-lives for each of the decay process.

Answer»

Solution :`""_(29)Cu""^(64)` UNDERGOES
(i) `B""^(-)` emission (ii) `B""^(+)` emission (iii) electron capture

Let the rate CONSTANT for `beta ""^(-)` emission = `K""_(1)`
Let the rate constant for `beta""^(+)`emission = `k""_(3)`
Let the rate constant for electron capture emission = `k""_(3)`
Overall rate constant
`K=K""_(1)+K""_(2)+K""_(3)=(0.693)/(t""_(1//2) )=(0.693)/(12.8)H""^(-1)`
`t""_(1//2)` of `""_(29)Cu""^(64)=12.8h`
`K""_(1)` = % of its decay `xx` Rate constant (overall)
`(38)/(100)xx k=(0.38xx0.693)/12.8h""^(-1)`
`t""_(1//2)` of `beta""^(-)` emission = `(0.693)/(K)=(0.693xx12.8)/(0.38xx0.693)`
`=(12.8)/(0.38)`
(partial half life) = 33.68 hr.
`K""_(2)` = % of `beta""^(oplus)` decay `xxK` (overall)
`=(19)/(10)xx(0.693)/(12.8)xx(0.19xx0.693)/(12.8)h""^(-1)`
Let `(t""_(2))""_(1//2)` be the half life of `beta""^oplus` emission
`=(0.693)/(K""_(2))=(0.693xx12.8)/(0.693xx0.19)=(12.8)/(0.19)`
= 67.37 hr
Discussion
91882.

A 0.05M KOHsolution offered resistanceof 31.6 omegain a conductivitycell of cel constant 0.3967 cm^(-1)at 298 k what is the molar conductance of KOHsolution

Answer»

150.3
18068
232
215.7

Solution :Givencellconstant `=0.367 cm^(-1)`
Resistance =31.6 `omega` and KOHis monoacidic base so itsmolarity =0.05 M
k(specific conuctance ) `= "CONDUCTANCE" xx "cell CONSTANT"`
`=(1)/("resistance")xx("cell constant")`
`=(0.367 cm^(-1))/(31.6 omega)=0.0116 omega^(-1) cm^(-1)`
`=0.0116 S CMS^(-1)`
`therefore` (molar conductance )`=(kxx1000)/(M)`
`=(kxx1000)/(0.05)=(0.0116xx1000)/(0.05)`
`=232 S cm^(2) mol^(-1)`
Discussion
91883.

""^(64)Cu (t_(½)=12.8h) decays by beta^(-) -emission (38%), beta^(+)-emission (19%) and electron capture (43%). Write decay products and calculate partial half-lives for each of the decay processes.

Answer»

SOLUTION :`lamda_(CU)= (0.6932)/(12.8)= 0.054`

`therefore lamda_(1)` = FRACTIONAL yield of `Zn xx lamda_(Cu)= (38)/(100) xx 0.054 = 0.0205`
`therefore t_(½)` for `beta^(-)` emission `=(0.6932)/(0.0205)= 33.8h`
Similarly we can CALCULATE, `t_(½)`for `beta^(+)` emission = 67.6hand `t_(½)` for ELECTRON capture= 29.85h
Discussion
91884.

A 0.025M solution of a monobasic acid had a freezing point of -0.06^(@)C. Calculate K_(a) for the acid . K_(f)(H_(2)O)=1.86

Answer»

Solution :`DeltaT_(f)` (observed)=`0.06^(@)C`, (f.p. of `H_(2)O=0^(@)C`)
`DeltaT_(f)` (calculated) `=0.025xx1.86=0.0465^(@)C`
(for DILUTE aqueous solution, molarity = molality)
`:.i=("observed depression in f.p.")/("calculated depression in f.p.")=(0.06)/(0.0465)=1.29`
For the acid HA we have
`{:(0.025,,,,0,,,,0,,"initial concentration"),(HA,,=,,H^(+),,+,,A^(-),,),(0.025(1-X),,,,0.025x,,,,0.025x,,"final concentration"):}`
(x is the degree of dissociation of HA)
`i=(1+x)/(1)=1.29`, `x=0.29`
`:.K_(a)=((0.25x)(0.025x))/(0.025(1-x))=0.025(x^(2))/(1-x)`
`=(0.025xx0.29^(2))/(1-0.29)=2.96xx10^(-3)`
Discussion
91885.

^64Cu(half - life = 12.8 h) decays by beta emission (38%) , beta^(+) emission (19%) and electron capture (43%) Write the decay products and calculate partial half - lives for each of the decay processes.

Answer»


ANSWER :N//A
Discussion
91886.

A 0.025 m solutio of mono basic acid had a freezing point of 0.06^(@)C. Calculate K_(s) for the acid K_(f) for H_(2)O=1.86^(@) "molality"^(-1).

Answer»

Solution :For mono basic acidHA
`HA hArr H^(+)A^(-)`
`{:(1,0,0),((1-alpha),alpha,alpha):}` "" i= (1+alpha)`
From Ostwald Dilution Law,`K_(a)=(CALPHA^(2))/((1-alpha))`
`DeltaT_(f)=K_(f)xx(i)`
`rArr 0.06=1.876xx0.025xx(1+alpha)`
`:.(1+alpha)=(0.06)/(0.0465)`
`:. alpha=0.029, C=0.025M`
`K_(s)=(alpha^(2))/((1-alpha))`
From equation (i) `K_(a)=(0.025xx(0.029)^(2))/((1-0.029))`
`=2.96xx10^(-3)`
Discussion
91887.

Nuclei of .^(64)Cu can decay be electron capture (probability 61%) or by beta^(+) decay (39%). Half life of .^(64)Cu is 12.7 hour. Find the partial half life for electron capture decay process,

Answer»


Solution :`beta^(-) "emission"`
`._(29)Cu^(64)rarr ._(30)ZN^(64) +._(-1)E^(0)`
`beta^(+) "emission"`
`._(29)Cu^(64) rarr ._(28)Ni^(64)+._(+1)e^(0)`
Electron capture
`.(29) Cu^(64)+_(-1)e^(0) rarr._(28)Ni^(64)`
`% "of I product" = 38%=(lambda_(1))/(lambda_(1)+lambda_(2)+lambda_(3)) xx 100 =(lambda_(1))/(LAMBDA)xx100=((0.693)/((t_(1//2))_(1)))/((0.693)/(t_(1//2)))xx100`
`lambda rArr` overall disintegration constant
`(38)/(100)=((t_(1//2)))/((t_(1//2))_(1))`
`:. (t_(1//2))=(t_(1//2))xx(100)/(38)=12.8 xx(100)/(38)=33.68 "hour"^(-1)`
similarly
`(t_(1//2))_(II)=(t_(1//2))xx(100)/(19)=12.8 xx(100)/(19)=67.36 "hour"^(-1)`
`(t_(1//2))_(III)=(t_(1//2))xx(100)/(43)=12.8xx(100)/(43)=29.776 "hour"^(-1)`
Discussion
91888.

6.488 g of lead combine directly with 1.002g of oxygen to form lead peroxide (PbO_(2)). Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxides 13.38 percent. Use these data to illustrate the law of constant composition.

Answer»

Solution :STEP 1. To calculate the PERCENTAGE of oxygen in first experiment.
Mass of peroxide formed `=6.488+1.002=7.490g`.
7.490 g of lead peroxide contain 1.002 g of oxygen
`THEREFORE 100G` of lead peroxide will contain oxygen `=(1.002)/(7.490)xx100=13.38g`, i.e., oxygen present = `13.38%`
Step. To compare the percentage of oxygen in both the experiments.
Percentage of oxygen in `PbO_(2)` in the first experiment = 13.38
Step 2. To compare the percentage of oxygen in both he experiments.
Percentage of oxygen in `PbO_(2)` in the second experiment = 13.38
Since the percentage composition of oxygen in both the SAMPLES of `PbO_(2)` is identical, the above data illustrate the law of constant composition.
Discussion
91889.

A 0.025-g sample of a compound that is composed of B and H, has a molecularmass of about 28 amu and burns spontaneously when exposed to air, producing 0.063 g of B_2O_3 . Find the molecular formula of the compound

Answer»

SOLUTION :`B_2H_6`
Discussion
91890.

632 g of sodium thiosulphate (Na_2S_2O_3) reacts with copper sulphate to form cupric thiosulphate which is reduced sodium thiosulphate to give cupros compound which is dissolved in excess of sodium thiosulphate to form a complex compound sodium cuprothiosulphate (Na_4[Cu_6(S_2O_3)_5]). CuSO_4+Na_2S_2O_3toCuS_2O_3+Na_2SO_4 2CuS_2O_3+Na_2S_2O_3toCu_2S_2O_3+Na_2S_4O_6 3Cu_2S_2O_3+2Na_2S_2O_3to underset("Sodium cuprothiosulphate")(Na_4[Cu_6(S_2O_3)_5]) In this process, 0.2 moles of sodium cuprothiosulphate is formed. (O=16, Na=23, S=32) The average oxidation states of sulphur in Na_2S_2O_3 and Na_2S_4O_6 are respectively.

Answer»

`+5 & +2`
`+2 & +2.5`
`+5 & 2.5`
`+2 & +4`

Solution :Suppose OXIDATION NUMBER of sulphur in `Na_2S_2O_3` is x
`2+2x+3(-2)=0`
or 2x=6-2
or 2x=4
x=+2
Suppose oxidation number of sulphur in `Na_2S_4O_6`is y
2+4Y+3(-2)=0
or 4y =12-2
or 4y=10
y=+2.5
Discussion
91891.

A 0.0200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced to copper (I) by iodine, 2Cu^(2+)+4I^(-)to2 CuI+I_2 If 20.0 mL of 0.10 M Na_2S_2O_3 is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu=63.5 g/mole)

Answer»

`31.75%`
`63.5%`
`53%`
`37%`

SOLUTION :2 MOLES of `Cu^(2+)`=1 moles of `I_2` =2 moles of hypo.
so moles of hypo used =`20xx10^(-3)xx0.1=2` MILLI moles = milli moles of COPPER
HENCE percentage of copper =`(2xx10^(-3)xx63.5)/0.2xx10%=63.5%`
Discussion
91892.

6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO_(2) evolved is absorbed completely in one litre of an aqueous solution of NaOH. A 10 mL portion of this solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. An another 10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl organe end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid and strength of original NaOH solution.

Answer»

Solution :`C_(x) H_(y) O_(2)+O_(2) rarr x CO_(2) overset("2 NaOH")(rarr) xNa_(2) CO_(3)`
` ( 6.4)/( M)``( 6.4x)/( M) ``( 6.4x)/( M)`
Let 100ml of solution contains a MEQ. Of NaOH & b meq. Of `Na_(2) CO_(3)`
when `HPh` is used as indicator then
meq. of NaOH `+ 1//2` meq. of `Na_(2) CO_(3) -` Meq. of HCl, `= ( 14.5xx 1)/( 10) xx 1000 = 40%`
`a + b//2 = 14.5 xx 100 = 1450`....(1)
when MeOH is used as indicator, meq. of `NaOH + `meq. of `Na_(2) CO_(3)` = Meq. of HCl
`a + b = 18 xx 1 xx(1)/( 10) xx 100 = 1800 `.....(2)
from ( 1) & ( 2)
`b = 700,a = 100,` moles of `Na_(2) CO_(3) = ( b xx 10^(-3))/( 2) = ( 6.4) /(M) xx x `
`rArr( 700 xx 10^(-3))/( 2) = 6.4 ( x )/(M).....(1) ` , wt% of =25
`M xx ( 25)/( 100) = 32rArr M = 128`
from ( 1) `( 700 xx 10^(-3))/( 2) = 6.4 xx ( x)/( 128)`
`( 700 xx 10^(-3))/( 2)xx ( 128)/( 6.4) = x = 7`
formula `C_(7) H_(12) O_(2)`
Molarity of NaOH `= ( 1800)/( 1000) = 1.8 M`
Discussion
91893.

A 0.016 M of an acid solution in benzene is dropped on a water surface, the benzene evaporates and the aci forms a monomolecular film of solid type. What volume of the above solution would be required to cover a 500 surface area of water with monomolecular layer of acid? Area covered by single acid molecule is 0.2

Answer»

`24.94xx10^(-3)ML`
`25.94xx10^(-3)ml`
`3.67xx10^(-3)ml`
`20.78xx10^(6)ml`

ANSWER :B
Discussion
91894.

6.4 gm of SO_(2) will contain how many (i) Moles of SO_(2) molecules (ii) Number of SO_(2) molecules

Answer»


ANSWER :`(##ALN_NC_CHM_MC_E01_005_A01##)`
(i) 0.2 MOLE(II)`(0.1xxN_(A))`
Discussion
91895.

A 0.010 M solution of PuO_2(NO_3)_2 was found to have a pH of 4.0. What is the hydrolysis constant, K_h", for "PuO_2^(2+),and what is K_b" for "PuO_2 OH^+ ?

Answer»


ANSWER :`10^(-6),10^(-8)`
Discussion
91896.

6.3 g of exalic acid (eq.wt=63) is dissoved in 500 mL. solution. What will be the normality of the solution ?

Answer»


Solution :`"Normalioty of solution (N)"=("Mass of oxalic acid/EQUIVALENT mass")/("VOLUME of solution in litres")`
`"Mass os oxalic acid = 6.3 g, Equivalent mass of oxalic acid "= 63" g equiv"^(-1)`,
`"Volume of solution=500 mL="500/1000=0.5 L`
`"Normality (N)"=((6.3g)/(63" g equiv"^(-1)))/((0.5L))=0.2"equiv L"^(-1)=0.2N.`
Discussion
91897.

A 0.01m aqueous solution of K_(3)[Fe(CN)_(6)] freezes ar -0.062^(@)C. What is the apparent percentage of dissociation? [K_(f) for water = 1.86]

Answer»

Solution :We have, `DeltaT_(f)=mxxK_(f)` …………(Eqn 7)
`=0.01xx1.86`
i.e., `(DeltaT_(f))_("normal")=0.0186^(@)`
and `(DeltaT_(f))_("observed")=0.062^(@)` (given)
`:.i=("observed COLLIGATIVE property")/("normal colligative property")`…………(Eqn. 9)
`=(0.062)/(0.0186)`
Now suppose x is the degree of DISSOCIATION of `K_(3)[Fe(CN)_(6)]`
Thus,
`{:("moles before DISS": :,, 1"MOLE",,0,,0),(K_(3)[Fe(CN)_(6)],,=,,3K^(+),,+[Fe(CN)_(6)]^(3-)),("moles after diss" : ,,(1-x),,3x,,x):}`
`:.i=((1-x)+3x+x)/(1)=(0.062)/(0.0186)`..........(Eqn.10)
`x=0.78`
`:.` per cent disscoiation `=78%`
Discussion
91898.

6.3 gram of oxalic acid , 0.4 mole of oxalic acid and 0.2 grams equivalents of oxalic acid are dissolved in 2 litre solution. 50ml of this solution neutralises. 20ml of KMnO_(4) solution in acidic medium. Molarity of KMnO_(4) solution is

Answer»

0.275M
0.550M
0.137M
0.350M

Answer :A
Discussion
91899.

A 0.02 M solution of a weak mono basic acid is 5% ionised. Calculate the ionisation constant of the acid.

Answer»

Solution :The degree of ionisation and the DISSOCIATION CONSTANT of the weak ACID are RELATED by the equation .
`K_a=(alpha^2 C)/(1-alpha)~= alpha^2 C`
`alpha=5% (or) 0.05`
`K_a=(0.05)^2 xx 0.02=0.00005`
`K_a=5xx10^(-5)`.
Discussion
91900.

61.8 g of A combine with 80 g of B. 30.9 g of A combine with 106.5 g of C. B and C combine to form compound CB_(2). Atomic weights of C and B are respectively 35.5 and 6.6. Show that the law of reciprocal proportions is obeyed.

Answer»

Solution :
MASS of B combining with 1 g of A `=(80)/(61.8)=1.26g`
Mass of C combining with 1 g of A `=(106.5)/(30.9)=3.45g`
Ratio of masses of B and C combining with FIXED mass of A `=1.29:3.45=1:2.67=3:8`
Ratio of masses of B and C combining DIRECTLY with each other `=13.2:35.3=1:2.67=3:8`
Thus, the two rations are same.
Discussion