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92251.

""^(36)Kr has the electronic configuration as [Ar]3d^(10)4s^(2)4p^(6). The next 37th electron willgo into the subshell

Answer»

4d
4f
5s
6s

Answer :C
Discussion
92252.

3.6gr of O_2 is adsorbed on 1.2gr of metal powder what volume of O_2 adsorbed per gram of the adsorbent at latm and 273K

Answer»

2.1 
0.19 
1
None 

SOLUTION :Mass of `O_2` per gram of ADSORBENT = `(3.6)/(1.2) = 3`
No. of MOLES of `O_2` pergram of adsorbent = `3/32`
Volume of `O_2` per gram of adsorbent `3/32 XX (0.0821 xx 273)/1 = 2.1`
Discussion
92253.

36g water and 828g ethyl alcohol form an ideal solution. The mole fraction of water in it, is

Answer»

`1.0`
`0.7`
`0.4`
`0.1`

Solution :`N=(W)/(M)=(828)/(46) = 18, n =(w)/(m)=(36)/(18)=2`
`x_(H_(2)O)=(n)/(n+N)=(2)/(2+18)=(2)/(20)=0.1`
Discussion
92254.

3.68g of mixture of CaCO_(3) and MgCO_(3) , on reaction with 1000 mL of N/10 HCl solution produced1.76" g of " CO_(2) . Calculate the percentage of each in the mixture .

Answer»

Solution :`CaCO_(3) +2HCI to CaCl_(2) +H_(2)O + CO_(2)`
`MgCO_(3) +2HCl to MgCl_(2) +H_(2)O + CO_(2)`
From the EQUATION , 2 eq. of HCl produces 1 MOLEOF `CO_(2)`
` :. `1 eq. HCl produces 1/2 molee of `CO_(2)`
` :." eq. wt of " CO_(2) = 44/2 = 22 `
` :. " eq. of " CO_(2) = (1.76 )/22 = 0.08"" ...(Eqn . 4i) `
From the GIVEN problem it is clear that the number of equivalents of the mixture of `caCO_(3) and MgCO_(3)` is lesser than that of HCLSOLUTION . Andso the `CO_(2)` is produced by the complete consuumption of `CaCO_(3) and MgCO_(3)` .
` :." eq. of " CaCO_(3) + " eq. of " MgCO_(3) = " eq. of " CO_(2)`
`x/50 +(3.68 - x)/42 = 0.08( " x = amount of " CaCO_(3))`
x = 2.0 g
` :. % of CaCO_(3) = 2/(3.68) xx 100 = 54 .34 % `
% of `Mg CO_(3) = 100 - 54. 34 = 45 . 66 % `
Discussion
92255.

36.4 g of 1,1,2,2-tetrachoropropane was heated with zinc dust and the product was bubbled through ammoniacal AgNO_(3). What is the mass of precipitatee obtained? Give equation for the reaction involved.

Answer»

Solution :
Mass of PROPYNE obtained from 36.4 G of
1,1,2,2-tetrachloropropane `=40/182 xx 36.4 = 8`g
`CH_(3)-C=-CH+AgNO_(3)to CH_(3)-C=-C-Ag + NH_(4)NO_(3)+H_(2)O`
Mass of precipate obtained from 8g of propyne
`=147/40 xx 8 = 29.4` g
Discussion
92256.

3.6 gram of oxygen is adsorbed in 1.2g of metal powder. What volume of oxygen adsorbed per gram of the adsorbent at STP ?

Answer»

`0.19Lg^(-1)`
`1 Lg^(-1)`
`2.1Lg^(-1)`
none of these

Solution :Mass of `O_(2)` per GRAM of adsorbent `=(3.6)/(1.2)=3`
No. of moles of `O_(2)` per gram of adsorbent `=(3)/(32)`
VOLUME of `O_(2)` per gram of adsorbent `=(3)/(32)XX(0.082.273)/(1)=2.10`
Discussion
92257.

36 g of a liquid ( molar mass =72 ) are dissolved in 72 g of H_(2)O to forman ideal solution. If the vapour pressure of pure water at a given temperature is 18 mm Hg, which one is the vapour pressure of water in the solution ?

Answer»

16 MM Hg
18 mm Hg
36 mm Hg
98 mm Hg

Answer :A
Discussion
92258.

3.55g of bleaching powder, when treated with acetic acid and excess of KI liberated Iodine which required 6 ml of 0.5 H Hypo solution. What is the percentage of available Cl_(2) ?

Answer»


Solution :% AVAILABLE chlorine `= (3.55 xx 0.5 xx V )/(W ) = ( 3.55 xx 0.5 xx 6)/( 3.55) = 3.0`
% available `Cl_(2) = 3%`
3.55g of bleaching POWDER, when treated with acetic acid and excess of KI liberated Iodine which required 6 ml of 0.5 N HYPO solution. What is the percentrage of available `Cl_(2)` ?
Discussion
92259.

3.55g of bleaching powder when treated with acetic acid and excess of KI Iiberatediodine which required 60mL of 0.6N hypo solution. Calculate the percentage of available chlorine.

Answer»

Solution :Number of milli EQUIVALENTS of HYPO, iodine and CHLORINE are same.
60ML of 0.6 N hypo solution =60 ml of 0.6 N `Cl_(2)` solution
Gram equivalent weight of chlorine -35.5g
percentage available chlorine=
`(60xx0.6xx35.5)/(1000xx3.55)xx100=36%`
Discussion
92260.

35.4 mL of HCl is required for the neutralization of a solution containing 0.275 g of sodium hydroxide. The normality of hydrochloric acid is

Answer»

0.97 N
0.142 N
0.194 N
0.244 N

Solution :We KNOW that 1 G EQUIVALENT weight of
NAOH = 40 g
`therefore` 40 g of NaOH = 1 g eq. of NaOH
`therefore 0.275 g` of `NaOH =(1)/(40xx0.275 eq.`
`=(1)/(40)xx0.275xx1000=6.88` meq.
`therefore underset((HCL))(N_(1)V_(1))=underset((NaOH))(N_(2)V_(2))`
`N_(1)xx35.4=6.88 "" (because meq = NV)`
`N_(1)=0.194`.
Discussion
92261.

(3.50 xx 10^2 mL) – (0.0225 L)may be written to correct significant digits :

Answer»

`3.28 XX 10^2 ML`
`0.3275 L`
`0.33L`

SOLUTION :`{:(3.50 xx 10^2 mL),(-0.225 xx 10^2 mL),(overline(3.275 xx 10^2 mL )):}`
` = 3.28 xx 10^2 mL`
(UPTO two decimal PLACES as in 0.50 )
Discussion
92262.

3.55 g sample of bleaching powder suspendedin H_(2)O was treated with enough acetic acid and KI solution. Iodine thus liberated required 80 mL of 0.2 M hypo for titration. Calculate th % of availablechlorine.

Answer»

SOLUTION :% of `Cl_(2)=(3.55xx0.2xx80)/(3.55)=16%`
Discussion
92263.

35 mL 0.1 M HCl solution is required to neutralise 25 mL. x M Ba(OH)_(2) solution. What is the molarity of Ba(OH)_(2) solution ?

Answer»

`0.14`
`0.28`
`0.35`
`0.07`

SOLUTION :`(M_(1)V_(1))/(1)=(M_(2)V_(2))/(2)`
`M_(1)xx25=(0.1xx35)/(2)=0.07`
Discussion
92264.

34.2 g of cansugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be

Answer»

<P>0.0099
1.597
0.84
0.9901

Solution :`(P_(1)^(@)-P)/P_(1)^(@)=(W_(2)//M_(2))/(W_(1)//M_(1)+W_(2)//M_(2))`
`=(34.2//342)/(34.2//342+180//18)=0.1/10.1=0.0099`
Discussion
92265.

34 g of hydrogen peroxide is present in 1120 ml of Solution. This solution is called

Answer»

10 VOLUME
20 volume
30 volume
32 volume

Solution :Wt. of `H_(2)O_(2) " in " 1 ml = 34/(1120) `G
` H_(2)O_(2) to H_(2)O + 1/2 O_(2)`
34 g of `H_(2) O_(2) ": gives " 11200 " MLOF " O_(2) ` at STP .
` :34/(1120) " g of " H_(2)O_(2) -= (11200)/34 xx 34/(1120) = 10 " ml of " O_(2) ` at STP .
Discussion
92266.

3,3-dimethylbutan-2-ol on treatment with conc. H_2SO_4 to give tetramethyl ethylene asa major product. Suggest a suitable mechanisms

Answer»

Solution :
According to Saytzeff.s rule the dehydration of 3,3-dimethylbutan-2-ol gives a MIXTURE of alkenes. But the SECONDARY carbocation formed in this reaction undergoes rearrangement to form a more STABLE tertiary carbocation which further, undergoes to B-elimination LEADS more stable product, that is 2,3-dimethyl but-2-ene (more YIELD).
Discussion
92267.

3,3-Dimethylbutan-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as the major product. Suggest a suitable mechanism.

Answer»

SOLUTION :
Discussion
92268.

3,3- dimethylbutan -2-ol on treatment with conc. H_(2)SO_(4) to give tetramethyl ethylene as a major product. Suggest a suitable mechanisms.

Answer»

SOLUTION :When 3,3- dimethyl-2-butanol is heated conc `H_(2)SO_(4)`, the ELIMINATION occurs after CARBOCATION REARRANGEMENT resulting in the formation of 2,3-dimethyl but-2-end.
Discussion
92269.

3,3-dimethyl butanoic acid is obtained from dry ice and what ?

Answer»

Iso-butyl-MgX
t-butyl-MgX
Neo-pentyl-MgX
t-pentyl-MgX

Answer :C
Discussion
92270.

(""_(32)Ge^(76), ""_(34)Se^(76)) and (""_(14)Si^(30), ""_(16)S^(32)) are example of

Answer»

isotopes and isobars
isobars and isotones
isotones and isotopes
isobars and isotopes.

Solution :`{:(""(32)^(76)Gerarr{:("protons = 32 = ELECTRONS"),("neutrons = 76 - 32 = 44"):}),(""(34)^(76)Serarr{:("protons = 34 = electrons"),("neutrons = 76 - 34 = 42"):}):}]{:("Mass no. same"),("but atomic no."),("different. So,"),("they are isobars."):}`
`{:(""(14)^(30)Sirarr{:("protons = 14 = electrons"),("neutrons = 30 - 14 = 42"):}),(""(16)^(32)Srarr{:("protons = 16 = electrons"),("neutrons = 32 - 16 = 16"):}):}]{:("Same no of"),("neutrons."),("So, they are"),("isotones."):}`
Discussion
92271.

3.28 g of a sample of pure copper when heated in presence of oxygen of some time forms black copper oxide (CuO) which weighs 3.92 g. What approximate percent of copper remains unoxidized?

Answer»

`4.6%`
`5.6%`
`6.6%`
`7.6%`

Solution :`{:(2Cu+O_(2)rarr,2CuO),(2xx6.5g,2(63.5+16)),("= 127 G"," = 159 g"):}`
3.92 g CuO will be obtained by oxidation of Cu
`=(127)/(159)xx3.92=3.13`
Cu LEFT unoxidized = 3.28 - 3.13 = 0.15 g
`THEREFORE %" of Cu left unoxidized"`
`=(0.15)/(3.28)xx100=4.57~=4.6`
Discussion
92272.

3.24 g of Hg(NO_(3))_(2) [M-Wt=324] dissolved in 1000g of water constitutes a solution having freezing point of =0.0558^(@)C while 21.68 "of" HgCl_(2) (molar mass =271) in 2000g of water constitutes a solution with a frezzing point of -0.0744^(@)C. The K_(f) for water is 1.86(K kg)/(mol) . About the state of ionisation of these two solids in water it can be inferred that

Answer»

`HG(NC_(3))_(2) and HgCl_(2)` both are COMPLETELY ionized
`Hg(NO_(3))_(2)` is fully ionized but `Hgcl_(2)` is fully unionized
`Hg(NO_(3))_(2)` and `HgCl_(2)` both are completely unionized
`Hg(NO_(3))_(2)` is fully unionized but `Hgcl_(2)` is fully ionized

ANSWER :B
Discussion
92273.

3.2 moles of hydrogen iodide were heated in a sealed bulb at 444^(@)C till the equilibrium state was reached. Its degree of dissociation at this temperaturee was found to be 22% The number of moles of hydrogen iodide present at equlibrium are

Answer»

`2.496`
`1.84`
2
4

Solution :`22/100xx3.2=0.704`
`THEREFORE"at equil. MOLES of"HI=3.2-0.704=2.496`
Discussion
92274.

""_(32) U^(238) (IIIB) undergoes following emissions : ""_(32)U^(238) overset( - alpha)(rarr) A overset( - alpha)(rarr) B overset( -beta)(rarr) C Which is // are correct statements

Answer»

A will be of IB group
A will be of IIIB group
B will be of IA group
C will be of IIIA group

Answer :B
Discussion
92275.

3.2 mole of hydrogen iodide were heated in a sealed bulb at444^Oc till the equilibriumn was reached . The degreee of dissociation of HI at this temperature was found to be 22% calculate the number of mole of hydrogen iodide present at equlibrium.

Answer»

2.496
1.87
2
4

Answer :A
Discussion
92276.

32 mg of pure ._(94)^(238)PuO_(2) has an activity of 6.4 xx 10^(7) sec^(-1) (i) What will be the half life of ._(94)^(238)Pu in years? (ii) What amount of PuO_(2) will remain if 100 mg PuO_(2) is kept for 500 years?

Answer»

Solution :(i) Mass of `.^(238)Pu = (238)/(270) xx 32 = 28.207 MG`
Rate `= (0.693)/(t_(1//)) xx (w)/("At.wt") xx N`
`6.4 xx 10^(7) = (0.693)/(t_(1//2)) xx (28.207 xx 10^(-3))/(238) xx 6.023 xx 10^(23)`
`t_(1//2)= 7.729 xx 10^(11) sec = 2.45 xx 10^(4)` YEARS
(ii), `(0.693)/(t_(1//2)) = (2.303)/(t) log_(10) ((N_(0))/(N))`
`(0.693)/(2.45 xx 10^(4)) = (2.303)/(5000) log_(10) ((100)/(N))`
`N = 86.7` mg
Discussion
92277.

32 grams of O_(2) at STP will occupy volume equal to

Answer»

22.4 LITERS
28 liters
22400 liters
2.24 liters

ANSWER :A
Discussion
92278.

3.2 g of oxygen (At wt. =16) and 0.2 g of hydrogen (At. Wt. =1) are placed in a 1.2 litre flask at 0^(@)C. The total pressure of the gas mixture will be.

Answer»

Solution :`3.2 g O_(2) =0.1` mol,
`0.2 g H_(2)=0.1` mol,
TOTAL `n=0.2 mol`,
`P=(nRT)/(V) =(0.2xx0.82xx273)/(1.12)=4` atm
Discussion
92279.

3.2 g oxygen is diffused in 10 minutes. In similar condition 2.8 g nitrogen will diffuse in :

Answer»

9.3 minutes
8.2 minutes
7.6 minutes
11.8 minutes

Answer :A
Discussion
92280.

32 g of oxygen and 3 g of hydrogen are mixed and kept in a vessel of 760 mm pressure and 0^@C. The total volume occupied by the maximum will be nearly :

Answer»

22.4 litre
33.6 litre
56 litre
44.8 litre

Answer :C
Discussion
92281.

3.2 g of oxygen and 0.2 g of hydrogen are placed in a 1.12 L flask at 0^(@)C. The total pressure of the gas mixture will be

Answer»

<P>1 ATM
4 atm
3 atm
2 atm

Solution :Moles of `O_(2) = ( 3.2 )/( 32) = 0.1`,
Moles of `H_(2) = ( 0.2) /( 2) = 0.1`
PRESSURE, `p = ( n RT )/( V )`
`= ( 0.2 xx 0.082 xx 273 )/( 1.12) = 4 atm`
Discussion
92282.

32 cc of hydrogen diffuses through a fine hole in 1 minute. What volume of CO_(2) will diffuse in 1 minute under the same conditions ?

Answer»

Solution :We have,
`(V_(1))/(V_(2)) = SQRT((M_(2))/(M_(1)))`
i.e.,`(V_(H_(2)))/(V_(CO_(2))) = sqrt((M_(CO_(2)))/(M_(H_(2))))`
`(32)/(V_(CO_(2)) = sqrt((44)/(2))`
`V_(CO_(2)) = 6.82` cc.
Discussion
92283.

3.17 g of a substance was deposited by the flow of 0.1 mole of electrons. The equivalent weight of the substance is

Answer»

3.17
0.317
317
31.7

Answer :D
Discussion
92284.

3.1 mole of FeCl_(3) and 3.2 mole of NH_(4)SCN are added to one litre of water. At equilibrium, 3.0 mol of FeSCN^(2+) are formed. The equilibrium constant K_(c) of the reaction : Fe^(3+)+SCN^(-)iffFeSCN^(2+) will be :

Answer»

`6.66 XX 10^(-3)`
0.3
3.3
150

Answer :B
Discussion
92285.

30g of magnesium and 30g of oxygen are reacted and the residual mixture contains

Answer»

45 g of magnesium OXIDE and 15G of OXYGEN
50g of magnesium oxide and 10g of oxygen
60g of magnesium oxide only
40g of magnesium oxide and 20g of oxygen

Solution :30 gram of Mg and 30 gram of `O_(2)` are reacted.
`2Mg+O_(2)=2MgO`
`2xx24` gram of magnesium combines with 32 gram of oxygen.
`therefore`30 gram of magnesium combines with
`(32xx30)/(2xx24)` gram of `O_(2)=20` gram of `O_(2)`.
Residual mixture contains (30-20) gram=10 gram
of `O_(2) and 50` gram of magnesium oxide.
Discussion
92286.

3.011xx10^(22) atoms of an element weighs 1:15gm. The atomic mass of the element is:

Answer»

23
10
16
35.5

Solution :ATOMIC MASS in gram=Mass of `6.022xx10^(23)` ATOMS
`=(1.15xx6.022xx10^(23))/(3.011xx10^(22))=23`
Discussion
92287.

300ml of "0.1M KMnO"_(4) in acid media is taken for oxidation of hydrogen peroxide. It is firstly diluted to 1 litre and then 50 ml of the solution is found to oxidize 40ml of H_(2)O_(2) to O_(2) . What will be the volume strength of H_(2)O_(2)?

Answer»

SOLUTION :m-eq of `KMnO_(4)` in 300 ml solution `= 300 xx0.5 = 150""7.5 =40xxN_(H_(2)O_(2))`
m-eq of `KMnO_(4)` in 1 litre diluted solution = 150`"or "N_(H_(2)O_(2))=0.1875N`
m-eq of `KMnO_(4)` in 50ML solution =` 7.5 = "m-eq. of "H_(2)O_(2)"Volume strength "=5.6xx0.1875="1.05 vol"`
Discussion
92288.

300 ml of gas at 27^(@)C is cooled to -3^(@)C at constant pressure. The final volume is

Answer»

350 ML
135 ml
270 ml
540 ml

ANSWER :C
Discussion
92289.

300 mL of gas at 27^@C is cooled to 10^@C at constant pressure, the final volume is :

Answer»

540 mL
135 mL
283 mL
350 mL

Answer :C
Discussion
92290.

300 mL mixture of ethylene and Butylenen is burnt in presence of oxgen then 150mL of oxygen is required, what is the volume of Ethylenen & Butylene in mixute.

Answer»

Solution :Let the volume of `C_(2)H_(4)=x mL`
So volume of Butylene `=(30-x)mL`
For `C_(2)H_(4)C_(2)H_4)+3O_(2)rarr2CO_(2)+2H_(2)O`
from equation
`because` for `1` volume `C_(2)H_(4). 3` volume of `O_(2)` is required.
`because` for `x mL` vol. of `C_(2)H_(4), 3x ml` volume of `O_(2)` is required.
For `C_(4)H_(8)``C_(4)H_(8)+6O_(2)rarr4CO_(2)+4H_(2)O`
`because` for `1` volume `C_(4)H_(8)' 6` volume of `O_(2)` is required.
Total volume of `O_(2)=3x+6(30-x)mL=150 mL` (Given)
`x=10`
`because` Volume of `C_(2)H_(4)` in mixture is `10ML`
`because` Volume of `C_(2)H_(4)` in mixture is `20mL`
Discussion
92291.

300 mL of a gas at 27^@ C and 700 mm pressure was brought under the conditions of 327^@C temperature and 140 mm pressure. What will be the volume ?

Answer»


ANSWER :300 ML
Discussion
92292.

300cm^(3) of an aqueous solution of a protein contains 2.12 g of the protein, theprotein, osmotic pressure of such a solution at 300 K is found to be 3.89xx10^(-3)" bar." Calculate the molar mass of the protein. ("R = 0.0823 L bar mol"^(-1)K^(-1))

Answer»

SOLUTION :`M_(2)=44852.44" G "MOL^(-1)`.
Discussion
92293.

30 volume hydrogen peroxide means:

Answer»

`30% H_2O_2` solution
`30 cm^3` of the solution contains 1 g of `H_2O_2`
`1 cm^3` of the solution liberates `30 cm^3` of `O_2` at STP
`30 cm^3` of the solution contains 1 MOLE of `H_2O_2`

ANSWER :C
Discussion
92294.

3.0 moles of PCl_5 kept in 1 L closed reaction vesselwas allowed to attain equilibrium at 380 K. The composition of the mixture i.e., PCl_5 and PCl_3 respectively at equilibrium is (K_c = 1.80)

Answer»

1.71,2.25
1.59, 1.61
1.41,1.59
1.41, 1.41

Solution :`{:(,PCl_(5(g)) hArr , PCl_(3(g)) + , Cl_(2(g))), ("Initial CONC. (mol/L)", 3.0 , 0,0),("At equilibrium" , 3-X, x ,x ) :}`
`K_c=x^2/(3-x)=1.8 RARR x=1.59`
Conc. Of `PCl_3` at equilibrium = 1.59
Conc. Of `PCl_5` at equilibrium = 3-1.59 = 1.41
Discussion
92295.

30 ml of sodium carbonate solution is mixed with 20 ml of 0.8N H_2 SO_4. The resultant solution needed 20 ml of 0.7 N HC1 for complete neutralization. The strength of Na_2 CO_3 solution will be:

Answer»

30 g/l
53 g/l
3.2 g/l
1.2 g/l

Answer :B
Discussion
92296.

30 mL of methanol (density 0.7980 g/mL) on mixing with 70 mL of water (density 0.9984 g/mL) at 298 K gave a solution of density 0.9575 g/mL. Calculate (i) mole fraction, (ii) molality, (iii) molarity, and (iv) f.p. of the solution. K_f(H_2O) = 1.86 .

Answer»

SOLUTION :0.1615, 10.7043, 7.6337, `-19.91^@C`
Discussion
92297.

30 ml of a solution containing 9.15 gm/litre of an oxalate K_(x)H_(y) (C_(2)O_(4))_(z) nH_(2)O are rerquired for titrating 27 ml of 0.12 N NaOH and 36 ml of 0.12 N KMnO_(4) separately. Calculate x,y,zand n. Assume all H atoms (except H_(2)O) are replaceable and x,y,z are in the simple ratio of gm atoms.

Answer»

Solution :Let,molecular weight of oxalate salt is M
(i) n-factor in ACID- base reaction = 2
(ii) n-factor in redox titration = `2 xx z`
`(C_(2)O_(4)^(2-) to 2CO_(2) + 2e)`
` :. ` Meq. Of acid in 30 ml = Meq.of NaOH used
`30 xx (9.15)/M xx y = 27 xx 0.12`
Also, `30 xx (9.15)/M xx (2z) = 36 xx 0.12`
From EQUATIONS (1) and (2)`y/(2z) = 27/36 rArry/z 3/2`
Also, total cationic CHARGE = total anionic charge
` :. x + y = 2z`
By equations (3) and (4)
` x : y : z :: 1 : 3 : 2` .
These are in SIMPLEST ratio and molecular formula is `KH_(3)(C_(2)O_(4))_(2). nH_(2)O`
Molecular weight of salt ` = 39 + 3 + 176 + 18n = 218 + 18n`
Form equation (1), M ` = (30 xx 9.15 xx 3)/(27 xx 0.12) = 254. 16 `
` :. 218 + 18n = 254.15`
` :. ` Oxalate salt is `KH_(3)(C_(2)O_(4))_(2) . 2H_(2)O`
Discussion
92298.

30 ml of 0.06 M solution of the protonated form of an anion of acid methonine (H_2A^(+)) is treated with 0.09 M NaOH. Calculate pH after addition of 20 ml of base pKa_1=2.28 and pKa_2=9.2

Answer»

5.5
5.74
9.5
None

Solution :`PH=(pK_(a_1)+pK_(a_2))/2 =(2.28+9.2)/2=5.74`
Discussion
92299.

30 g of urea (M=60 g mol^(-1)) is dissolved in 864 g of water. Calculater the vapour pressure of water for this soution if vapourpressure of pure waer at 298 K is 23.8 mm Hg.

Answer»


SOLUTION :ACCORDING to RAOULT's Law
`(P_(A)^(@)-P_(S))/P_(S)=W_(B)/M_(B)xxM_(A)/W_(A)`
`((23.8mm)-P_(S))/P_(S)=((30g)xx(18" G mol"^(-1)))/((60" g mol"^(-1))xx(846 g))=0.0106`
`(23.8 mm)-P_(S)=0.02106xxP_(S)`
`(23.8 mm)=1.0106xxP_(S)`
`P_(S)=((23.8 mm))/(1.0106)=23.55mm`
Discussion
92300.

30 g of urea (M = "60 g mol"^(-1)) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.

Answer»

<P>

SOLUTION :`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))"(solution is TAKEN as dilute as it is less then 5%)"`
`(23.8-p_(s))/(23.8)=(30//60)/(846//18)"or"p_(s)=23.55mm`
Discussion