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93451.

0.44 g of a colourless oxide of nitrogen occupies 224 ml at STP. The compound is :

Answer»

`N_2O`
NO
`N_2O_2`
`NO_2`

ANSWER :A
Discussion
93452.

0.436g of acetyl derivative of a polyhydric alcohol (molecular mass =92) required 0.336g KOH for hydrolysis. Calculate the number of hydroxy gropus in the alcohol.

Answer»

Solution :We known that , `(w)/(w) (M+42n)=56 N`
where w is the amount of alkali, W amount ofacetyl DERIVATIVE, M is the molecular mass of alcohol and n NUMBER of HYDROXYL groups.
`(0.336)/(0.436)(92+42n)=56n`
or `92+42n=(0.436)/(0.336)xx56n=(218)/(3)n`
or `276+126n=218n`
or `n=3 ` ( The alcohol is trihydric. )
Discussion
93453.

0.42g of an organic compound containing C, H, O and N gave on combustion 0.924g of CO_(2) and 0.243g of water. 0.208 g of the substance when distilled with NaOH gave NH_(3), which required 30 mL of (N/20) H_(2)SO_(4) solution for neutralization. Calculate the amount of each element in 0.42g of the compound.

Answer»

SOLUTION :Carbon Moles of C in `CO_(2)=1 xx` moles of `CO_(2)`
`=1 xx (0.924)/(44)= 0.021`
Weight of `C= 0.021 xx 12= 0.252g`
HYDROGEN Moles of H in `H_(2)O = 2 xx` moles of `H_(2)O`
`=2 xx (0.243)/(18)= 0.027`
Weight of `H= 0.027 xx 1= 0.027g`
NITROGEN m.e. of `H_(2)SO_(4)= (1)/(20) xx 30= 1.5`
`therefore` m.e. of `NH_(3)= 1.5`
Eq of `NH_(3)= (1.5)/(1000)= 0.0015`
moles of `NH_(3)= 0.0015`
Now moles of N in `NH_(3)=1 xx` moles of `NH_(3)= 0.0015`
Weight of N in 0.208g of the compound = `0.0015 xx 14`
=0.021g
`therefore` weight of N in 0.42g of the compound `=(0.021)/(0.208) xx 0.42` = 0.042g
Oxygen `therefore` TOTAL weight of C, H and `N= (0.252 + 0.027 + 0.042)g`
=0.321g
`therefore`weight of `O= 0.42- 0.321`
= 0.099g
Discussion
93454.

0.400 g of an acid HA (mol. mass = 80) was dissolved in 100 g of water. The solution showed a depression of freezing point of 0.12 K. What will be the dissociation constant (in multiple of 10^(-3) ) of the acid at about 0°C? Given K_f (water) = 1.86 K Kg "mol"^(-1)(Assume molarity of solution a molality)

Answer»


Solution :MOLALITY (m) of solution `=(0.4 xx 1000)/(80 x 100) = 0.05`
`DeltaT_(f)` (normal) `=K_(f) xx m = 1.86 xx 0.05 = 0.093` K
Van.t Hoff factor,
`i=(DeltaT_(f) ("OBSERVED"))/(DeltaT_(f)("normal")) = 0.12/0.093 = 1.290`
`HA + H_(2)O `ALPHA =(i-1)/(n-1) = (1.290-1)/(2-1) = 0.29`
`K_(a) = (Calpha^(2))/(1-alpha) = (0.05 xx (0.29)^(2))/(1-0.29) = 5.92 xx 10^(-3)`
Discussion
93455.

0.4 gm of He in a bulb at a temperature of 'T' K had a pressure of 'P' atm. When the bulb was immersed in hotter bath at a temperature 50K more than the first one, 0.08 gm of gas had to be removed to restore the original pressure. Then value of 'T' is:

Answer»

100K
200K
300K
500K

Solution :Since P and V is CONSTANT
`n_(i)T_(i)=n_(F)T_(f)`
`(0.4)/(4)xx(T)=((0.4-0.08))/(4)xx(T+50)`
`5T=4(T+50)`
T=200K
Discussion
93456.

0.4 g of NaOH present in one litre solution shows the pH:

Answer»

12
2
6
10

Answer :A
Discussion
93457.

0.3g of a compound on combustion gave 0.54g of water and 0.88g of carbon dioxide. Find the percentage of carbon and hydrogen in the compound

Answer»


ANSWER :80%, 20%
Discussion
93458.

0.396 gm of a bromoderivative of a hydrocarbon (A) when vapourized occupied 67.2 ml at NTP. On reaction with aqueous NaOH, (A) gives (B). (B) when passed over alumina at 250^(@)C gives a neutral compound (C ), while at 350^(@)C it gives a hydrocarbon (D): (D) when heated with HBr gives an isomer of (A). When (D) is treat with conc. H_(2)SO_(4) and the product is diluted with water and distilled, (E) is obtained. Identify (A) to (E) and explain the reaction involved.

Answer»

Solution :Molecular WEIGHT of `A=(0.369xx22.4xx1000)/(67.2)=123`
(A) is a bromoderivative, so it may be written as R - Br.
`M_(RBr)=123`, So, R = 43 `because` Br = 80 (Atomic weight)
i.e. `C_(n)H_(2n+1)=43`
or `12n+2n+1=43`
or n = 3
`THEREFORE A=C_(3)H_(7)Br`
So, the POSSIBLE structures of (A) are `CH_(3)CH_(2)CH_(2)BrorCH_(3)-UNDERSET(Br)underset(|)(CH)-CH_(3)`
The reactions are as follow.

Since (D) gives an isomer of (A) on HBr addition, hence
(A) = `CH_(3)CH_(2)CH_(2)Brand(E)=CH_(3)underset(OH)underset(|)(CH)CH_(3)`
Discussion
93459.

0.395g of an organic compound by Carius method for the estimation of S gave 0.582g of BaSO_(4). The percentage of S in the compound is :

Answer»

`20.24%`
`35.62%`
`12.24%`
`40.65%`

Solution :`underset(233)(BaSO_(4))-=underset(32)(S)`
`233g` of `BaSO_(4)` contain SULPHUR `=32g`
`0.582g` of `BaSO_(4)` contain sulphur
`=(32)/(233)xx0.582`
Percentage of sulphur
`=(32xx0.582)/(233xx0.395)xx100=20.24%`
Discussion
93460.

0.38 gm of a silver salt of a dibasic acid on ignition gave 0.27 gm of silver. Molecular mass of acid is x xx 10 gm then x value is

Answer»


Solution :`E=("weight of SILVER salt")/("weight of silver") xx 108 -107`
Discussion
93461.

0.38 g sample of NaNO_3is dissolved in 250 mlflask. The molarity of the solution is

Answer»

0.018 M
0.095 M
0.260 M
0.016 M

Solution :Moles of `NaNO_3 = (0.38)/(85)`
MOLARITY`= ( (0.38)/(85) XX 1000)/(250) = 0.018 M `
Discussion
93462.

0.36g of Mg combines with chlorine to produce 1.425 g of mangesium chlorine 9.50g of another sample of anhydrous magnesium chloride gave, on electrolysis 2.24 litre of chlroine at NTP. Show that these data agree with the law of constant proportions.

Answer»

SOLUTION :Mass of 2.24 LITRE of CHLORINE at NTP`=(71)/(22.4)xx2.24=7.1`
Discussion
93463.

0.36g of an organic compound containing sulphur produced H_(2)SO_(4) by Carius method, which on treatment with BaCl_(2) produced quantitatively 0.2330g of BaSO_(4). Calculate the percentage of S in the compound. (Ba= 137, S= 32, O= 16)

Answer»

SOLUTION :Moles of S in `BaSO_(4) = 1 XX` moles of `BaSO_(4)`
`=(0.2330)/(233)= 0.001 (BaSO_(4)= 233)`
WEIGHT of `S= 0.001 xx 32G`
`=0.032g`
% of `S= (0.032)/(0.36) xx 100= 8.89%`
Discussion
93464.

0.369g of a bromo derivative of a hydrocarbon (A) when vaporised occupied 67.2mL at NTP. (A), on reaction with aqueous NaOH, gives (B). (B), when passed over alumina at 250^(@)C, gives a neutral compound (C ) while at 300^(@)C, it gives a hydrocarbon (D). (D), when heated with HBr gives an isomer of (A). When (D ) is treated with concentrated H_(2)SO_(4) and the product is diluted with water and distilled, (E ) is obtained Identify (A) to (E ).

Answer»

SOLUTION :
Discussion
93465.

0.3605 g of a metal is deposited on the electrode by passing 1.2 ampere current for 15 minutes through its salt. Atomic weight of the metal is 96. what will be its valency?

Answer»


SOLUTION :CALCULATE the WT. deposited by 96500 coulombs. This will be the equivalent wt.
Then valency=At. Wt/Eq. wt.\
Discussion
93466.

0.32g of a metal gave on treatment with an acid 112mL. Of hydrogen at NTP. Calculate equivalent mass of the metal:

Answer»

58
32
11.2
24

Solution :MASS of metal tthat DISPLACES 11200ml, HYDROGEN at STP will be its equivalent mass.
`therefore` Equivalent mass of metal`=(0.32)/(112)xx11200=32`
Discussion
93467.

0.32 g of a metal on treatment with an acid gave 112 mL of hydrogen at NTP. Calculate the equivalent weight of the metal

Answer»

58
32
11.2
24

Solution :Equivalent WEIGHT of a metal is the mass of the metal which produces 11200 mL of `H_(2)` at NTP.
112 ml of `H_(2)` at NTP are PRODUCED from metal
`=0.32 g`
`therefore 11200` mL of `H_(2)` will be produced from metal
`=(0.32)/(112)xx11200`
= 32 g
Discussion
93468.

0.31 gm of an ally Fe+ Cu was dissolved in excen dilute H_(2)SO_(4) and the solution was made up to100 ml . 20 ml of this soltin required 3 m of N/30 K_(2)Cr_(2)O_(7) solution for exact oxidation. The % purity (in closest value) of Fe in wire is ____________.

Answer»


ANSWER :9
Discussion
93469.

0.30g of an organic compound contaiining C, H and Oxygen an combustion yields 0.44gCO_2 and 0.18g H_2O. If onemol of compound weight 60, then molecular formula of the compound is

Answer»

`C_2H_4O_2`
`C_4H_6O`
`C_3H_8O`
`CH_2O`

SOLUTION :Percentage of `C = 12/44 XX 0.44/0.30 xx 100 = 40%`
Percentage of `H= 12/18 xx 0.18/0.30 xx 100 = 6.6%`
Percentage of `O = 100-(40 + 6.6) = 53.4%`

Hence empirical formula `= CH_2O`
`n = ("Molecular mass")/("Emperical formula mass") =60/30 = 2`
`rArr ` Molecular formula of the compound` = (CH_2O)_2 = C_2H_4O_2`
Discussion
93470.

0.30g of an organic compound containing C, Hand Oxygen on combustion yields 0.44g CO_(2) and 0.18g H_(2)O. If one mole of compound weighs 60, then molecular formula of the compound is

Answer»

`C_(3)H_(8)O`
`C_(2)H_(4)O_(2)`
`CH_(2)O`
`C_(4)H_(6)O`

Solution :PERCENTAGE of `C=(12)/(44)XX(0.44)/(0.30)xx100=40%`
Percentage of `=(2)/(18)xx(0.18)/(0.30)xx100=6.6%`
percentate of `O=100-(40+6.6)=53.4%`

Hence, EMPIRICAL formula `=CH_(2)O`
`n=("Molecular mass")/("Empirical formula mass")=(60)/(30)=2`
`rArr` Molecular formula of COMPOUND `=(CH_(2)O)_(2)`
`=C_(2)H_(4)O_(2)`
Discussion
93471.

0.3 g of an acid is neutralised by 40 cm^(3) of 0.125 N NaOH. Equivalent mass of the acid is

Answer»

60
45
30
63

Solution :`40cm^(3)` of 0.125 N NAOH neutralise ACID = 0.3 g
`therefore 1000cm^(3)` of 1 N NaOH will neutralise acid
`=(0.3)/(40)xx(1000)/(0.125)g=60g`
`1000cm^(3)` of 1 N NaOH contain one g ram equivalent of NaOH. It will neutralise one gram equivalent of the acid.
Hence, equivalent WEIGHT of acid = 60
Discussion
93472.

02g of a monobasic organic acid containing C, H and O on combustion gave 0.505g of CO_(2) and 0.0892 g of H_(2)O. 0.183g of this acid required 15cc of N/10 NaOH for exact neutralisation. Find the molecular formula of the acid.

Answer»

Solution :Moles of `C=1 xx` moles of `CO_(2)= (0.505)/(44)=0.0115`
Wt. of `C= 0.0115 xx 12= 0.1380g`
Moles of `H=2 xx` moles of `H_(2)O=2 xx (0.0892)/(18)=0.0099`
wt of `H= 0.0099 xx1= 0.0099g`
Wt of O = wt of the compound -(wt. of C + wt. of H)
`=0.2 -(0.1380 + 0.0099)= 0.0521g`
Moles of `O= (0.0521)/(16)=0.0033`
`THEREFORE` moles of `C: H: O= 0.0115: 0.0099: 0.0033`
`=115: 99: 33`
`=3.5:3:1=7:6:2`
Empirical formula is `C_(7)H_(6)O_(2)`
Now, since the acid is monobasic, mol wt. = eq. wt
`therefore` EQUIVALENT of the acid `=(0.183)/("eq.wt")=(0.183)/("mol wt")`
`therefore` m.e. of the acid `=(0.183)/("mol. wt") xx 1000`
`=(183)/("mol.wt")`
m.e. of the BASE `=(1)/(10) xx 15=1.5`
`because` m.e. of the acid= m.e. of the base `(183)/("mol wt")=1.5 therefore` mol. wt.=122
Since the empirical formula weight, i.e., `84+6+32+122`, is equal to the molecular formula weight, the molecular formula of the acid is the same as the empirical formula. Hence the formula is `C_(7)H_(6)O_(2)`.
Discussion
93473.

0.2964g of Cu was deposited on passage of a current of 0.5 amp for 30 min through a solution of copper sulphate. Calculate the atomic weight of Cu.

Answer»


ANSWER :63.56
Discussion
93474.

0.2864 of Cu was deposited on passage of a current of 0.5 ampere for 30 minutes through a solution of copper sulphate. What is the electrochemical equivalent of copper ? "(1 F = 96500 coulombs)"

Answer»

Solution :Electrochemical equivalent (Z) is defined as the weight of the substance deposited by the passage of 1 coulomb of electricity.
`therefore z = ("wt. of Cu deposited")/("CHARGE in COULOMBS")`
`= ("wt. of Cu")/("current in AMPERE " xx " time in seconds")`
`= (0.2864)/(0.5 xx 30 xx 60)`
= 0.00032g/coulomb.
Discussion
93475.

0.28 g of nitrogenous was subjected to Kjeeldahl's proicess to produce 0.17 g of NH_(3). The percentage of nitrogen in the organic compound id :

Answer»

5
10
50
80

Solution :N//A
Discussion
93476.

0.2614g of an organic compound gave upon combustion 0.2324g of CO_(2) and 0.0950g of H_(2)O. 0.1195g of this compound gave 0.3470g of AgCl. If the vapour density of the substance is 49.5, calculate its molecular formula.

Answer»

SOLUTION :`(C_(2)H_(4)Cl_(2))`
Discussion
93477.

0.261 g ofa sample of pyrolusite was heated with excess of HCl and the chlorine evolved was passed in a solution of Kl . The liberated iodine requuired 90 mL N/30Na_(2)S_(2)O_(3). Calculate the percentage of MnO_(2) in the sample.

Answer»


SOLUTION :N/A
Discussion
93478.

0.25g of an organic compound on Kjeldahl's analysis gave enough ammonia to just neutralise 10cm^(3) of 0.5M H_(2)SO_(4). The percentage of nitrogen in the compound is

Answer»

`28`
`56`
`14`
`112`

Answer :B
Discussion
93479.

0.2595g of an organic compound yielded quantitatively 0.35g of BaSO4. Find the percentage of sulphur in the compound. (Ba= 137.3, S= 32, O= 16)

Answer»


ANSWER :`18.5%`
Discussion
93480.

0.2595 g of an organic substance in a quantitative analysis yielded 0.35 g of the barium sulphate. The percentage of sulphur in the subtance is

Answer»

`18.52 G`
`182.2 g`
`17.5 g`
`175.2 g`

Solution :`%` of `S = 32/233 XX (" w.t of " BaSO_(4))/(" w.t of ORGANIC COMPOUND") xx 100`
`= 32/233 xx (0.35)/(0.2595) xx 100 = 18.52% gm`.
Discussion
93481.

0.25 mole of propane is subjected to combustion. If this reaction is used for making a fuel cell, the number of moles of electrons involved in each half cell for this amount of propane will be

Answer»

Solution :No of moles of electrons involved for 1 MOLE of `C_(3)H_(8)=20`
`therefore`. No. of moles of electrons involved for 0.25 mole of `C_(3)H_(8)=5`.
Discussion
93482.

0.25 g of an element 'M' reacts with excess fluorine to produce 0.547g of the hexafluoride MF_(6). What is the element?gt

Answer»

Cr
Mo
S
Te

Answer :B
Discussion
93483.

0.2475g of an organic compound on combustion gave 0.4950g of CO_(2) and 0.2025g of H_(2)O. Calculate the percentage of oxygen in the compound

Answer»


ANSWER :0.3637
Discussion
93484.

0.245 g of nitrogenous organic compound was Kjeldahlised and ammonia was absorbed in 10 ml of 0.1 N H_(2)SO_(4). The unreacted acid requires 3ml of 0.1N NaOH for complete neutralization. Calculate the percentage of nitrogen in the organic compound.

Answer»


SOLUTION :By Kjeldhal METHOD
Discussion
93485.

0.2435 g of a complex gave 0.2870 g of AgCl when treated with a excess AgNO_(3) solution. The complex is

Answer»

`[Cr(NH_(3))_(4)Cl_(2)]Cl`
`[Cr(NH_(3))_(5)Cl_(2)]Cl_(2)`
`[Cr(NH_(3))_(3)Cl_(3)]`
`[Cr(NH_(3))_(6)Cl_(2)]Cl_(3)`

Solution :No. of moles of AGCL = `(0.287)/(143.5) = 0.002` moles
B can give 2 moles of AgCl.
Discussion
93486.

0.220g of a sample of a volatile compound, containing carbon, hydrogen and chlorine yielded on combustion in oxygen 0.195g of CO_(2), 0.0804g of H_(2)O. 0.120g of the compound occupied a volume of 37.24mL at 105^(@)C and 768mm Hg pressure. Calculate the molecular formula of the compound.

Answer»

Solution :Moles of C in `CO_(2)=1 xx` moles of `CO_(2)= (0.195)/(44)=0.00443`
Weight of `C= 0.00443 xx 12=0.05316g`
Moles of H in `H_(2)O =2xx` moles of `H_(2)O= 2 xx (0.0804)/(18)= 0.00893`
Weight of `H= 0.00893 xx1= 0.00893g`
`therefore` weight of Cl= 0.22 `-(0.05316 + 0.00893)g= 0.15791g`
Moles of `Cl= (0.15791)/(35.5)= 0.00445`
Moles of `C: H: Cl= 0.00443: 0.00893: 0.00445`
`=443: 893: 445=1:2:1`
i.e., the empirical formula is `CH_(2)Cl`
Now, volume of the vapour of 0.12g of compound at NTP
`=(37.24 xx 768)/(378)xx (273)/(760)=27.18mL`
Moles of the compound `=(0.12)/(M)` (M = mol .wt)
`=(27.18)/(22400)`
`thereforeM = 99`
`therefore ("molecular formula weight")/("empirical formula weight")= (99)/(49.5)=2`
Hence, the empirical formula is `(CH_(2)Cl)_(2)`, i.e. `C_(2)H_(40Cl_(2)`
Discussion
93487.

0.22 g sample of volatile compound, containing C, H and Cl only on combustion in O_2 gave 0.195 g CO_2 and 0.0804 g H2O. If 0.120 g of the compound occupied a volume of 37.24 mL at 105^@ and 768 mm of pressure, calculate molecular formula of compound.

Answer»

SOLUTION :`C_2H_4Cl_2`
Discussion
93488.

0.2060 g of a substance gave 18.8mL of moist nitrogen at 17^(@)C and 756mm pressure. If the vapour tension at 17^(@)C is 14.5mm, find the percentage of nitrogen in the compound

Answer»


ANSWER :0.1056
Discussion
93489.

0.21 g of an organic compound containing C, H, O and N gave upon combustion 0.462g of CO_(2) and 0.1215g of H_(2)O. The ammonia produced on distillation of 0.104g of this compound with NaOH, required 15mL of N/20 H_(2)SO_(4) for neutralisation. Find the empirical formula of the compound

Answer»

SOLUTION :`(C_(7)H_(9)NO_(2))`
Discussion
93490.

0.2033 g of an organic compound in Dumas method gave 31.7m L of moist nitrogen at 14^(@)C and 758mm pressure. Calculate the percentage of nitrogen in the compound. (Aqueous tension at 14^(@)C=14mm, N= 14)

Answer»

Solution :Pressure due to nitrogen only `=758-14=744mm`
VOLUME of nitrogen at NTP `=(744 xx 31.7 xx 273)/(287 xx 760)`
=29.52mL
Mole of nitrogen `=(29.52)/(22400)`
Weight of nitrogen `(N_(2))= (29.52)/(22400) xx 28= 0.0369g`
% of nitrogen `=(0.0369)/(0.2033) xx 100`= 18.16%
Discussion
93491.

0.20 g of a hydrocarbon on combustion gave 0.66 g CO_2 . The percentage of hydrogen in the hydrocarbon is about :

Answer»

33
45
10
90

Answer :C
Discussion
93492.

0.2 percent solution of phenol is a/an

Answer»

ANTISEPTIC
DISINFECTANT
antibiotic
antifertility drug

Answer :A
Discussion
93493.

0.2% solution of Phenol is an :

Answer»

ANTIBIOTIC
ANTISEPTIC
Disinfectant
Analgesic

Answer :B
Discussion
93494.

0.2 mole sample of hydrocarbon C_(x)H_(y) yields after complete combustion with excess O_(2) gas, 0.8 mole of CO_(2), 1.1 mole of H_(2)O Hence, hydrocarbon is

Answer»

`C_(4)H_(10)`
`C_(4)H_(6)`
`C_(4)H_(5)`
`C_(8)H_(16)`

ANSWER :A
Discussion
93495.

0.2 molar solution of formic acid is ionized to 3.2 %. Its ionisation constant is :

Answer»

`9.6xx10^(-3)`
`2.05xx10^(-4)`
`1.25xx10^(-6)`
`4.8xx10^(-5)`

SOLUTION :`K_(a)=c alpha^(2), c = 0.2 M, alpha = (3.2)/(100)=0.032`
`K_(a)=0.2xx(0.032)^(2)=2.05xx10^(-4)`
Discussion
93496.

0.2 molar solution of formic acid is ionized 3.2%. Its ionizationconstant is

Answer»

`9.6 XX 10^(-3)`
`2.1 xx 10^(-4)`
`1.25 xx 10^(-6)`
`4.8 xx 10^(-5)`

SOLUTION :`K_(a) = C ALPHA^(2) = 0.2 xx ((3.2)/(100))^(2) = 2.048 xx 10^(-4) = 2.1 xx 10^(-4)`.
Discussion
93497.

0.2 molal acid HX is 20% ionised in solution K_"f" = 1.86 K "molality"^(–1). The freezing point of the solution is -

Answer»

– 0.45
– 0.90
– 0.30
–0.53

Answer :A
Discussion
93498.

0.2 M NaOH is titrated with 100 ml 0.2 M CH_(3) COOH in a conducitivity cell. The data obtained were plotted as following The pH of solution in conductivity cell is 9 at point B. Calculate pH of conductivity cell when 100 ml 0.1 M HCl added in resulting solution present in conducitivity cell at point B.

Answer»


Solution :`CH_(3)COOH(AQ)+NaOH(aq)rarrCH_(3)COONA(g)+H_(2)O`
`{:(,(0.2xx100)/(1000),(0.2xx100)/(1000)," "0," "0),(," "0,0,0.02," "0.02):}`
`pH=9=(1)/(2)(pKw+pKa+log .(0.02)/(0.2))`
` 18=14+pKa-1`
`pKa=5`
`CH_(3)COONa(aq)+HCl(aq)rarrCH_(3)COOH(aq)+NaCl(aq)`
`{:(,0.02,(100xx0.1)/(1000)=0.01,0,0),(,0.01," "0,0.01,0.01):}`
Acidic buffer
`pH=pKa+log.(CB)/(CA)`
`pH=5+log.(0.01)/(0.01)`
`ph=5`
Discussion
93499.

0.2maqueoussolutionof KC1freezes at - 0.680^(@)C. Calculatevan'tHofffactorand osmoticpressureof solutionat0. ^(@)C (K_(f)= 1 .86 Kkg "mol"^(-1))

Answer»

SOLUTION :van'tHofffactor=`1= 1.83 , pi_(OB) = 8.19 ATM`
Discussion
93500.

0.2 m aqueous solution of a weak acid (HX) is 20% dissociated. The boiling point of this solution is ( K_(b) for water = 0.52 Km^(-1))

Answer»

`101.04^(@)C`
`100.104^(@)C`
`100.1248^(@)C`
`100.52^(@)C`

Solution :`{:("HX(AQ)",rarr,"H"^(+),+,"X"^(-)),(m(1-0.20),,m(0.20),,m(0.20)):}`
MOLAL CONC. of soln.
`=m(1-0.20)+0.20m +0.20m`
`=1.20m`
`DeltaT_(b)=K_(b) xx m`
`=0.52 xx 1.20(0.2)=0.1248`
Boiling point `=100.1248^(@)C`
Discussion