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92051.

50 ml of a solution, containing 0.01 mole each Na_(2)CO_(3),NaHCO_(3) and NaOH was titrated with N-HCl. What will be the titre readings if MeOH is added after the first end point with Ph

Answer»


ANSWER :20ML
Discussion
92052.

50 mLof a solution , containing 1 g each of Na_(2)CO_(3) ,NaHCO_(3) and NaOHwas titrated with N HCl . What will be the titre readings if only phenophthalein is used as indicator ?

Answer»

Solution :The titration reactions in this case are
`Na_(2)CO_(3) +HCl to NaHCO_(3) +NaCl`
and `NAOH +HCl to NaCl +H_(2)O `
Thus , we have
m.e of `Na_(2)CO_(3) + " m.e of NaOH = m.e of "v_(1) mL (say ) of N HCl `
`1/106 xx1000 +1/40 XX 1000 = 1 xx v_(1) :. v_(1) = 34.4 mL `
Discussion
92053.

50 ml of a solution, containing 0.01 mole each Na_(2)CO_(3),NaHCO_(3) and NaOH was titrated with N-HCl. What will be the titre readings if only MeOH is used as indicator from the beginning .

Answer»


ANSWER :40ML
Discussion
92054.

50 mL of an aqueous solution of glucose C_(6)H_(12)O_(6) (Molar mass : 180 g/mol) contains 6.02xx10^(22) molecules. The concentration of the solution will be

Answer»

0.1 M
0.2 M
1.0 M
2.0 M

ANSWER :A
Discussion
92055.

50 mL of a gas A diffuse through a membrane in the same time as for the diffusion of 40 mL of gas B under identical conditions of pressure and temperature. If the molecular mass of A is 128, that of B wouldbe :

Answer»

200
500
400
160

Answer :A
Discussion
92056.

50 ml of a gaseous hydrocarbon (having 85.7% carbon) at STP on combustion with oxygen gave 0.370 g of CO_2, 0.142 g H_2 O and an undetermined quantity of methane. The molecule formula of hydrocarbon will be:

Answer»

`C_4 H_8`
`C_2 H_2`
`C_5 H_8`
`C_2 H_4`

ANSWER :A
Discussion
92057.

50 ml of a solution, containing 0.01 mole each Na_(2)CO_(3),NaHCO_(3) and NaOH was titrated with N-HCl. What will be the titre readings if only Ph is used as indicator.

Answer»


ANSWER :20ML
Discussion
92058.

50 mL of a 0.1 M CuSO_(4) solution is electrolysed for 12 minutes at a current of 0.06 amp. If Cu is produced at one electrode and O_(2) at the other, what will be the pH of the final solution ? For HSO_(4)^(-) = H^(+) + SO_(4)^(2-), K_("diss") = 1.3 xx 10^(-2)

Answer»


ANSWER :2.95
Discussion
92059.

50 mL of 10 N H_(2)SO_(4) , 25 mL of 12 N HCl and 40 mL of 5 N HNO_(3)were mixed together and the volume of the mixture was made 1000 mL by adding water. The normality of the resultant solution will be

Answer»

1 N
2 N
3 N
4 N

Solution :`N_(1)V_(1) + N_(2)V_(2) + N_(3)V_(3) = NV`
`10 xx 50 + 12 xx 25 + 5 xx 40 = N xx 1000` or, N = 1N
Discussion
92060.

50 ml of 2 N acetic acid mixed with 10 ml of 1N sodium acetate solution will have an approximate ph of (K_a=10^(-5))

Answer»

4
5
6
7

Answer :A
Discussion
92061.

50 mL of 1 M oxalic acid (molecule mass = 126) is shaken with 0.5 g of oxalic acid adsorbed per gram gram of chrocoal.

Answer»

Solution :CALCULATION of initial amount of oxalic acid in 50 mL solution. 1 M oxalic acid solution means 1 mole of oxalic acid (126 g) present in 1000 mL solution.
`THEREFORE` 50 mL of 1 M solution will CONTAIN oxalic acid `=(126)/(1000)xx50=6.6g`
Calculation of amount of oxalic acid in 50 mL solution after adsorption. Conentration of solution after adsorption `=0.5M.` Thus, 1000 mL of the solution contain oxalic acid `=(63)/(100)xx50=3.15g`
Calculation of amount adsorbed per gram
Amount of oxalic acid adsorbed by `0.5` g charcoal `=6.30-3.15g =3.15g`
`therefore` Amount of oxalic acid adsorbed per gram of charcoal `=6.30g`
Discussion
92062.

50 ml of 1 M oxalic acid is shaken with 0.5 g wood charcoal. The final concentration of the solution after adsorption is 0.5 M. What is the amount of oxalic acid adsorbed per gram of carbon?

Answer»

3.15 G
3.45 g
6.30 g
NONE of these.

Solution :Amountof oxalicacid intitallypresent`= (1 xx 50)/(1000) xx 63 = 3.15g `
Amountof oxalicacidafter adsorption.
` = (50)/(1000) xx 63 xx 0.5 = 1.575 g`
Amountadsorbed` = 3.15 - 1.575 = 1.575 g`
Amountadsrobedper GRAM of charocoal`= 1.575 xx 2= 3.15 g`
Discussion
92063.

50 mL of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding 50 mL of NaOH solution. The remaining titration is completed by adding 0.5 N KOH solution. What is the volume of KOH solution required for completing the titration?

Answer»

Solution :CALCULATION of HCL left after 1st titration :
`{:(underset((HCl))(N_(1)xxV_(1))=underset((NaOH))(N_(2)xxV_(2)),,),(0.2xxV_(1)=0.1xx50,or,V_(1)=25mL):}`
i.e., 25 mL of 0.2 N HCl has been consumed and therefore 0.2 N HCl left `=50-25=25mL`
Calculation of KOH used in 2nd titration :
`{:(underset((KOH))(N_(1)xxV_(1))=underset((HCl))(N_(2)xxV_(2)),,),(0.5xxV_(1)=0.2xx25,or,V_(1)=10mL):}`
`therefore KOH` solution required for completing the titration = 10 mL.
Discussion
92064.

50 ml of 1 M acetic acid is skaken with 0.5 g wood charcoal. The final conc. of the solution after adsorption is 0.5 M. What is the amount of acetic acid adsorbed per gram of carbon?

Answer»


SOLUTION :`M = (wt)/(GMT) cdot (1000)/V , 1 = (wt)/(60) cdot 1000/50 " wt= 3 after " 0.5 = (wt)/(60) cdot (1000)/(50) wt = 15`
adsorbed = `(3 - 1.5) = 1.5 = 0.5 gr to 1.5 {:(1 to ?),(0,3):}`
Discussion
92065.

50 ml of 0.2 M solution of a compound with empirical formula CoCl_(3).4NH_(3) on treatment with excess of AgNO_(3)(aq) yields 1.435 g of AgCl. Ammonia is not removed by treatment with concentrated H_(2)SO_(4). The formula of the compound is :

Answer»

`Co(NH_(3))_(4)Cl_(3)`
`[Co(NH_(3))_(4)Cl_(2)]Cl`
`[Co(NH_(3))_(4)]Cl_(3)`
`[CoCl_(3)(NH_(3))]NH_(3)`

Solution :`n_(AGCL)=1.435/143.5=0.01`
=10 m MOL,
m mol of compound =50`xx`0.2=10
`rArr` Only 1 `Cl^(-)` permolecule is ionisable
Discussion
92066.

50 ml of 0.1 N HCl and 50 ml of 0.1 N NaOH are mixed. The pH of the resulting solution is :

Answer»

1
2
7
9

Answer :C
Discussion
92067.

50 ml of 0.1 M HCl and 50 ml of 0.2 M NaOH are mixed. The pH of the resulting solution is :

Answer»

`1.30`
`4.2`
`12.70`
`11.70`

Solution :`[H^(+)]` in 50 ml of 0.1 M HCl
`=(0.1xx50)/(1000)=0.005 M`
`[OH^(-)]`in 50 ml of 0.2 M NaOH
`=(0.2xx50)/(1000)=0.010 M`
EXCESS of `OH^(-)` ions present after neutralisation
`=0.010-0.005=0.005`
Concentration of `OH^(-)`
`=(0.005)/(100)xx1000=0.05 M`
`pOH=-log[OH^(-)]=-log0.05=1.30`
Discussion
92068.

50 mL of 0.1 M CuSO_(4) solution is electrolysed using Pt electrodes with a current of 0.965 ampere for a period of 1 minute. Assuming that volume of solution does not change during electrolysis, calculate [Cu^(2+)], [H^(+)] and [SO_(4)^(2-)] after electrolusis. What will be the concentration of each species if current is passed using Cu electrodes ?

Answer»


ANSWER :`[CU^(2+)] = 0.094M, [H^(+)] = 0.012 M, [SO_(4)^(2-)] = 0.1 M ;`
Discussion
92069.

50 ml of 0.05 M HNO_3is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.

Answer»

SOLUTION :Number of moles of `HNO_3 =0.05xx50xx10^-3`
`=2.5xx10^-3`
Number of moles of KOH `=0.025 xx50xx10^(-3)`
`=1.25xx10^-3`
number of moles of `HNO_3` after mixing
`=2.5xx10^(-3)-1.5xx10^(-3)`
`=1.25xx10^(-3)`
`therefore` concentration of `HNO_3=("Number of moles of"HNO_3)/("Volume is LITRE")`
After mixing, total volume `= 1000 ml =100xx10^(-3)L`
`therefore [H^+]=(1.25xx10^(-3)"moles")/(100xx10^(-3)L)`
`=1.25xx10^2` moles `L^(-1)`
`pH=-log[H^+]`
`pH=-log(1.25xx10^(-2))=2-0.0969`
`=1.9031`.
Discussion
92070.

50 ml of 0.05 M HNO_(3)is added to 50 mlof 0.024 M KOH. Calculate the pH of theresultantsolution

Answer»

Solution :Numberofmolesof ` HNO_(3) = 0.05 XX 50 xx 10^(-3)`
`=2.5 xx 10^(-3)`
Number of MOLES of KOH = ` 0.025 xx 50 xx 10^(-3)`
` 1.25 xx 10^(-3)`
Number of moles of ` HNO_(3)`after mixing
2.5 xx 10^(-3) - 1.5 xx 10^(-3)`
`1.25 xx 10^(-3)`
CONCETRATION of ` HNO_(3)`
` = ( " Number of molesof ` NHO_(3)) /( " Volume of littre") `
After mixing , totalvolume =` 100 ml = 100 xx 10^(-3) L `
` [ H^(+)] = ( 1.25 xx10^(-3) " moles")/( 100 xx 10^(-3) L)`
` 1.25 xx 10^(-2) " moles " L^(-1)`
PH= `log [ H^(+)]`
` pH= - log( 1.25 xx 10^(-2)) = 2- 0.0969`
` = 1.9031 `
Discussion
92071.

50 ml of 0.05 M HNO_3 is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.

Answer»

SOLUTION :Number of MOLES of `HNO_3=0.05 times 50 times 10^-3=2.5 times 10^-3`
Number of moles of KOH=`0.025 times 50 times 10^-3=1.25 times 10^-3`
Number of moles of `HNO_3` after mixing =`2.5 times 10^-3-1.5 times 10^-3=1.25 times 10^-3`
`therefore " Concentration of " HNO_3= ("Number of moles of" HNO_3)/( "Volume in litre")`
After mixing total volume=`1.25 times 10^-2moles L^-1`
`therefore[H^+]=(1.25 times10^-3 MOL es)/(100 times10^-3L)=1.25 times10^-2mol es L^-1`
`pH=-log[H^+]`
`pH=-log(1.25 times10^-2)=2-0.0969`
`=1.9031`
Discussion
92072.

50 mL 10 N H_(2)SO_(4), 25 ml 12 N HCl and 40 ml 5 N HNO_(3) were mixed together and the volume of the mixture was made 1000 ml by adding water. The normality of the resultant solution will be:

Answer»

1 N
2 N
3 N
4 N

Solution :`N_(1)V_(2)+N_(2)V_(2)+N_(3)V_(3)=N_(4)V_(4)`
`10xx50+12xx25+5xx40=N_(4)xx1000`
`"or"N_(4)=1N`
Discussion
92073.

50 ml 10N-H_(2)SO_(4),25ml12N-HCl and 40 ml" "5N-HNO_(3) were mixed together and the volume of the mixture wasmade 1000 ml by adding water. The normality of the resultant solution will be

Answer»

1N
2N
3N
4N

Solution :`H_(2)SO_(4)" HCl "HNO_(3)` Total volume
`N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=Nxx1000ml`
`N=(N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3))/(1000)`
`=(50xx10+25xx12+40xx5)/(1000)`
`N=(500+300+200)/(1000)=(1000)/(1000)=1N`
Discussion
92074.

50 gm of CaCO_(3) is allowed to react with 68.6 gm of H_(2)PO_(4) then select the correct option(s)- 3CaCO_(3)+ 2H_(3)PO_(4) to Ca_(3)(PO_(4))_(2) + 3H_(2)O + 3CO_(2)

Answer»

51.67 gm salt is formed
Amount of unreacted REAGENT =35.93 gm
`n_(CO_(2))`=0.5 MOLES
0.7 moles `CO_(2)` is evolved

Answer :A::B::C
Discussion
92075.

5.0 g of marble was added to 7.5 g dilute hydrochloric acid. After the reaction was over, it was found that 0.5 g of marble was left unused. Calculate the percentage strength of hydrochloric acid. What volume of CO_(2) measured at STP will be evolved in the above reaction?

Answer»


Solution :`underset(100g)(CaCO_(3))+underset(2xx36.5g=73g)(2HCl)rarrCaCl_(2)+H_(2)O+underset(22400cm^(3)" at STP")(CO_(2))`
`"MARBLE reacted "=5-0.5 G=4.5g`
`"HCl reacted with 4.5 g marble"=(73)/(100)xx4.5g=3.285g`
`%" strength"=(3.285)/(7.5)xx100=43.8%`
`CO_(2)" evolved at S.T.P."=(22400)/(100)xx4.5cm^(3)=1008cm^(3)`
Discussion
92076.

50 g of ethylene glycol is dissolved in 170.3 g of water at -9.3^(@)C . The amount of water seperated as ice is ________g.

Answer»


ANSWER :9
Discussion
92077.

50 drops each of water and ether, weigh 3.64 g and 0.852 g respectively. Determine the surface tension of ether if the surface tension of water is 72.75 dyne cm^(-1)

Answer»


ANSWER :17.03 DYNE `CM^(-1)`
Discussion
92078.

50% completion of a first order reaction takes place in 16 minutes. Then fraction that would react in 32 minutes from the beginning

Answer»

`1//2`
`1//4`
`1//8`
`3//4`

ANSWER :D
Discussion
92079.

"50 cm"^(3) of 0.2" N HCl" is titrated against 0.1" N NaOH"The remaining titration adding "50 cm"^(3) of NaOH. The remaining titration is completed by adding 0.5 N KOH The volume of KOH required for completing the titration is

Answer»

A) `"12 cm"^(3)`
B) `"10 cm"^(3)`
C) `"25 cm"^(3)`
D) `10.5" cm"^(3)`

SOLUTION :Milli - equivqlents of `HCl=50xx0.2=10` milli-eq.
Milli-equivalents of `NaOH=50xx0.1=5` milli - eq.
Remaining milli - equivalents of acid = 5
REQUIRED volume of `KOH=0.5xxV_(1)=5`
`V_(1)=10cm^(3)`
Discussion
92080.

50 cm^3of 0.2N HCl is titrated against 0.1NNaOH solution. The titration is discontinued after adding 50 cm^3of NaOH. The remaining titration is completed by adding 0.5N KOH. The volume of KOH required for completing the titration is :

Answer»

10 `cm^(3)`
12 `cm^(3)`
16.2 `cm^(3)`
21.0 `cm^(3)`

Solution :We know,1 milliequivalent`=NxxV` (in ML)
Milliequivalents of HCl `=0.2xx50=10`
Milliequivalents of NaOH=`0.1xx50=5`
`therefore` HCl and NaOH neutralise each other with equal equivalents
M.eq. of HCl left=10-5=5
Volume of new solution =50+50=100`cm^(3)`
`N_(HCl)"left"=(5)/(100)=0.05` N (from above formula)
`100 cm^(3)` of 0.05 N titrated with 0.5 N KOH with volume,
`N_(HCl)V_(HCl)=NKOHV_(KOH)`
`V=(0.05xx100)/(0.5)=10cm(3)`
Discussion
92081.

50 cm^(3) of 0.04 M K_2Cr_2O_7 in acidic medium oxidizes a sample of H_2S gas to sulphur. Volume of 0.03M KMnO_4 required to oxidize the same amount of H_2S gas to sulphur, in acidic medium is

Answer»

`60cm^(3)`
`80cm^(3)`
`90cm^(3)`
`120cm^(3)`

SOLUTION :`K_(2)Cr_(2)O_(7)` and `KMnO_(4)` both are good oxidizing agents in acidic medium. The reaction is as follows :
`Cr_(2)O_(7)^(2-) + 14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O`
`MnO_(7)^(2-)+8H^(+)+5e^(-) rarr MN^(2+)+4H_(2)O`
Normality of `Cr_(2)O_(7)^(2-)=` MOLARITY `xx` no. of ELECTRONS involved
`= 0.04xx6= 0.24` N
Normality of `MnO_(4)^(-)=` Molarity `xx` no. of electrons involved
`=0.03xx5= 0.15` N
`underset((K_(2)Cr_(2)O_(7)))(N_(1)V_(1))= underset((KMnO_(4)))(N_(2)V_(2))`
`rArr0.24xx50=0.15xxV_(2)`
`V_(2)=(0.24xx50)/(0.15)=80cm^(3)`
Discussion
92082.

5% (wt./vol.) aqueous NaCl solution and 5% (wt./vol.) aqueous NaCl solution are:

Answer»

Isotonic
Isomolar
Isoequimolar
None

Answer :D
Discussion
92083.

5% (wt./vol.) aqueous NaCl solution and 5% (wt./vol.) aqueousKCl solution are:

Answer»

Isotonic
Isomolar
Isoequimolar
None

Answer :D
Discussion
92084.

5 Write structures of the products of the following reactions: (i)CH_(3) - CH= CH_(2) overset (H_(2)O //H^(+)) to (ii)(##NCERT_CHE_XII_C11_E01_005_Q01.png" width="80%"> (iii) CH_(3) - CH_(2) - underset (CH_(3)) underset (|)(CH)- CHOoverset(NaBH_(4)) to

Answer»

SOLUTION :(i) `CH_(3) - UNDERSET (OH) underset (|) (CH) - CH_(3)`
(ii)
(iii) `CH_(3) - CH_(2) - underset (OH) underset (|) (CH) - CH_(2)OH `
Discussion
92085.

5% solution of a substnace in water has freezing point 269.06K. Calculate molar mass of solute. Freezing point of pure water 273.15K. [K_(f)=14K.kg" "mol^(-1)].

Answer»

SOLUTION :`180G" "MOL^(-1)`.
Discussion
92086.

5% sucrose solution is isotonic with 1% 'X' solution. What will be the molecular weight of 'X' ?

Answer»

343 gram/mole
180 gram/mole
68.4 gram/mole
171 gram/mole

SOLUTION :For the 5 % solution `(5)/(M_(1))` and 1% solution `= (1)/(M_(2))`
`therefore (5)/(342)=(1)/(M_(2))`
`therefore M_(2)=68.4` gram/mole.
Discussion
92087.

5-oxohexanal is obtained by ozonolysis of :

Answer»




ANSWER :B
Discussion
92088.

5 moles of SO_(2) and 5 moles of O_(2) in a closed vessel. At the equlibrium stage 60% of SO_(2) is used up. The total number of moles of SO_(2), O_(2)and SO_(3) in the vessel now is

Answer»

`10.0`
`8.5`
`10.5`
`3.9`

ANSWER :B
Discussion
92089.

5 moles of SO_(2) and 5 moles of O_(2) are allowed to react to form SO_(3) in a closed vessel. At the equilibrium stage, 60%SO_(2) is used up. The total number of moles of SO_(2),O_(2)andSO_(3) in the vessel now is

Answer»

`3.9`
`10.5`
`8.5`
`10.0`

Solution :`{:(,2SO_(2),+,O_(2),iff,2SO_(3)),("INITIAL moles",5,,5,,),("At eqm.",5-(60)/(100)xx5,,5-1.5,,3),(,=5-3=2,,=3.5,,):}`
Total = `2+3.5+3=8.5`
Discussion
92090.

5 moles of SO_(2) and 5 moles of O_(2) are allowed to react to form SO_(3) in a closed vessel. At the equilibrium stage 60% of SO_(2) is used up. The total number of moles of SO_(2), O_(2) and SO_(3) in the vessel now is

Answer»

10.5
3.9
`10.0`
8.5

Solution :`{(2SO_(2),+,O_(2),hArr,2SO_(3)),("Initially" rarr 5 "mole",,"5 mole",,"0 mole"),(" At eq" rarr 5(1-0.6),,5(1-0.3),,5 XX 0.6),("mole",,"Mole",,"Mole"):}`
When 60% `SO_(2)` is used up, the `O_(2)` used up = 30% & `SO_(3)` formed is 60% of initial AMOUNT of `SO_(2) rArr` TOTAL no. of mole `=2 + 3.5 + 3 = 8.5`
Discussion
92091.

5 moles of SO_(2) and 5 moles of O_(2) are allowed to react, At equilibrium, it was found that 60% of SO_(2) is used up. If the partial pressure of the equilibrium mixture is one atmosphere, the partial pressure of O_(2) is

Answer»

<P>0.52 atm
0.21 atm
0.41 atm
0.82 atm

Solution :`2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`
Initially `5""5""0`
Degree of dissociation =0.60
`:.` no. of moles of `SO_(2)` at equilibrium =2
no. of moles of `O_(2)` at equilibrium =3.5
no. of moles of `SO_(3)` produced at equilibrium =3
`:.p_(O_(2))=(3.5xx1)/8.5=0.41` atm
Discussion
92092.

5 moles of H_(2)SO_(4) contain

Answer»

5 eq. of H
5 moles of S
20 eq. of O
10 eq. each of H, S and O

Answer :B::D
Discussion
92093.

5 moles of CaCO_(3) on heating yeilded 2 moles of CO_(2) .Find % yield of reaction.

Answer»


ANSWER : `(##ALN_NC_CHM_MC_E01_026_A01##)`
`40%`
Discussion
92094.

5 moles of AB_(2) weigh 125xx10^(-3) kg and 10 moles of A_(2)B_(2) weigh 300xx10^(-3) kg. The molar mass of A(M_(A)) and molar mass of B(M_(B) in kg mol_(-1) are :

Answer»

`M_(A) = 50xx10^(-3) and M_(B) = 25xx10^(-3)`
`M_(A) = 25xx10^(-3) andM_(B) = 50xx10^(-3)`
`M_(A) = 5xx10^(-3) and M_(B) = 10xx10^(-3)`
`M_(A) = 10xx10^(-3) andM_(B) = 5xx10^(-3)`

ANSWER :C
Discussion
92095.

5 mole of SO_(2)and 5 moles of O_(2) are allowed to react. At equlibrium, it was found that 60% of SO_(2) is used up. If the partial pressure of the equlibrium mixture is one atmosphere, the partial pressure of O_(2) is

Answer»

`0.52` ATM
`0.21` atm
`0.41` atm
`0.82` atm

Solution :Initially `underset(5)(2SO_(2(G)))+underset(5)(O_(2(g)))hArrunderset(0)(2SO_(3(g)))`
Degree of dissociation `=0.60`
`therefore` no of MOLES of `SO_(2)` at equlibrium =2
no of moles of `O_(2)` atequlibrium =3.5
no of moles of `SO_(3)` PRODUCED at equlibrium=3
`therefore pO_(2)=(3.5xx1)/(8.5)=0.41atm.`
Discussion
92096.

5 ml of N HCl, 20 ml of N/2 H_(2)SO_(4) and 30 ml of N/3 HNO_(3) are mixed together and the volume made to one litre. The normality of the resulting solution is

Answer»

N/5
N/10
N/20
N/40

Solution :`N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(4)V_(4)=N_(4)(V_(1)+V_(2)+V_(3))`
Discussion
92097.

5 mL of acetone is mixed with 100mL of H_(2)O. The vapour pressure of water above the solution is

Answer»

equal to the vapour pressure of pure WATER
equal to the vapour pressure of the SOLUTION
less than the vapour prssure of pure water
more than tha vapour pressure of pure water

Solution :V.P. of any COMPONENT in the solution = mole fraction of the componentx V.P. of that component in the pure STATE and mole fraction is always lt 1.
Discussion
92098.

5 mLof 8 N nitricacid , 4.8 mLof 5 HCl and a certain volume of 17M sulphuric acid are mixed togther and made up to 2 litres . 30 mL of this mixture exactly neutralises 42.9 mL of sodium carbonate solution containing 1 g of Na_(2)CO_(3) . 10 H_(2)O in 100 mL of water . Calculate the amount in grams of the sulphate ions on the solution .

Answer»

Solution :Let the volume of 17 M ( i.e ., 34 N) `H_(2)SO_(4)` solution be v mL .
` :. ` total m.e of the acid mixture = `8 xx 5 + 5 xx 4.8 + 34 v "" ..(Eqn.1)`
= `(64 + 34v)`
` :.` normal of the mixture = `(m.e)/( " total volume (mL)") ""…(Eqn .1)`
` = (64 + 34 v)/(2000) `
m.e of 30 mL ofthis acidmixture` = (64+ 34v)/2000 xx 30`
Now , NORMALITY of `Na_(2)CO_(3) . 10 H_(2)O ` solution = `(g//litre)/("eq.wt")`
` = 10/43 `
` {{:( "grams/litre of "Na_(2)CO_(3).10H_(2)O=10),("and eq. wt"=(mol.wt)/2 = 286 /2 ):}}`
` :. ` m.eof 42.9 mL of `Na_(2)CO_(3).10H_(2)O ` solution = `10/143 xx 42.9`
THUS , m.e of 30 mL of acid mixture
= m.e of `42.9 " mL of " Na_(2)CO_(3) . 10 H_(2)O ` solution .
` :. ` m.eof N . (i.e ., 17 M) `H_(2)SO_(4) = 34 xx 68/17 "" ...(Eqn.1)`
` = 136 `
` :.` equivalent of `H_(2)SO_(4) = 136/1000 = 0.136 "" ...(Eqn.3)`
` :. " equivalent of"SO_(4)^(2-) = 0.136 ""....(Eqn . 7II)`
weight of `SO_(4)^(2-)= eq xx eq .wt of SO_(4)^(2-) "" ....(Eqn.4i)`
` = 0.136 xx 48 `
` = 6.528 g `
`( " eq. wt of " SO_(4)^(2-)= (" ionix wt")/("valency")= 96/2 = 48 )`
Discussion
92099.

5mL of N HCI, 20 mL of N//2 H_(2)SO_(4) and 30 mL of N//3 HNO_(3) are mixed together and volume made to one litre. The normality of the resulting solution is

Answer»

N/5
N/10
N/20
N/40

Solution :`N_(1)V_(1)+N_(2)V_(2)+N_(3)V_(3)=N_(4)V_(4)`
`1xx5+1//2xx20_//3xx30=N_(4)xx1000`
`5+10+10=N_(4)xx1000`
`N_(4)=25/1000=1/40`
Discussion
92100.

5 mL of 0.4N NaOH is mixed with 20 mL of 0.1N HCl. The pH of the resulting solution will be :

Answer»

6
7
8
5

Solution :`{:(NaOH,+,HCL rarr NaCl+H_(2)O),(5mL,,20ML),(0.4N,,0.1N):}`
Moles of `NaOH=(0.4)/(1000)xx5=2xx10^(-3)` moles
Moles of `HCl = (0.1)/(1000)xx20=2xx10^(-3)` moles
As the moles of NaOH and HCl are EQUAL, the resulting solution will be NEUTRAL.
Discussion