Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

An aqueous solution containing one mole per litre of each Cu(NO_(3))_(2),AgNO_(3),Hg_(2)(NO_(3))_(2) and Mg(NO_(3))_(2) is being electrolysed using inert electrodes. The values of the standard electrode potentials in volts (reduction potentials) are Ag|Ag^(+)=0.80,2Hg|Hg_(2)^(2+)=+0.79 Cu|Cu^(2+)=+0.34,Mg|Mg^(2+)=-2.37 With increasing voltage, the sequence of deposition of metals on cathode will be

Answer»

AG,HG,Cu,Mg
Mg,Cu,Hg,Ag
Ag,Hg,Cu
Cu,Hg,Ag

Solution :HIGHER the reduction potential, more EASILY is the metal deposited `(M^(n+)+n e^(-)toM)`. `Mg^(2+)` ions will not be deposited because `H_(2)O` is reduced more easily than `Mg^(2+)` ions.
2.

An aqueous solution containing urea found to have boiling point more than the normal boiling point of water (373.13 K). When the same solution was cooled, it was found that its freezing point is less than the normal freezing point of watre (273.13K). Explain there observations.

Answer»

Solution :In aqueous solution, urea ACTION as a non-volatile solute with very little vapour pressure of its own. However, it OCCUPIES somesurgace AREA. Since evapouration is a surface phenomerna, the vapour pressure of the solutio gets reduced. This means that the solution has to be heated to a higher temperature so that the vapour pressure on its surface may match the atmospheric pressure that corresponds to the biling point temperature of the solution. In other words, the boiling point of the solution gets RAISE. For more detailos consult Section 10.
Reverse happents to the freezing point of the solution when teh same non-volatile solute is added to it. A solution freezes when the vapour pressure on its surface becomes equal to that of pure solvent. On adding a non-volalite solute, the vapour when the vapour pressure of the solution becomes equal to that of TH pure sovent at a lower temperature. In other wards, freezing point temperature of dthe solution gets lowerd. For more details, consults Section 11.
3.

An aqueous solution containing S^(2-) ions will not give:

Answer»

A YELLOW precipitate with the suspension of `CdCO_(3)` in water
Black precipitate with lead ACETATE solution
White precipitate with `CaCO_(3)` suspension
Purple colour with sodium thiosulphate solution

Answer :C::D
4.

An aqueous solution containing one mole per litre of each Cu(NO_3)_2 , AgNO_3 . Hg(NO_3)_(2), Mg(NO_3)_2 is being electrolysed using inert electrodes. The values of standard electrode potential (reduction potential) in volts are Ag//Ag^(+) =+ 0.80 V ""Hg//Hg^(2+) = + 0.79V "" Cu//Cu^(2+) = +0.34 V "" Mg//Mg^(2+) = -2.37 V With increasing voltage, the sequence of deposition of metals on cathode will be

Answer»

Ag , Hg , CU ,Mg
Mg, Cu, Hg, Ag
Cu, Hg, Ag
Cu, Hg, Ag, Mg

Answer :C
5.

An aqueous solution containing one mole per litre of Cu(NO_3)_2 , AgNO_(3),Hg(NO_3)_2 and Mg(NO_3)_2 is being electrolysed using inert electrodes. The values of the standard electrode potentials in volts (reduction potientials) are Ag |Ag^(+) = +0.80,2Hg|Hg^(2+)=0.79 Cu|Cu^(2+) = + 0.34, Mg|Mg^(2+) =-2.37 With increasing voltage, the sequence of deposition of metals on cathode will be

Answer»

Ag,HG,Cu,Mg
Mg,Cu,Hg,Ag
Ag,Hg,Cu
Cu,Hg,Ag

Solution :`rho = R.(a)/(l) = (25 xx 3.2)/(1.6) =50 ` ohm cm
`K = (1)/(rho) = (1)/(50) =0.02 S cm^(-1)`
`Lambda_(eq) = K xx V = K xx (1000)/("Normality")= (0.02 xx 1000)/(0.5)`
`=40 s cm^2 ` EQUIV.
6.

An aqueous solution containing one gram of urea boils at 100.30^(@)C. The aqueous solution containing 3.0 g of glucose in the same volume will boil at

Answer»

`100.90^(@)C`
`100.60^(@)C`
`100^(@)C`
`100.30^(@)C`

SOLUTION :MOLAR CONCENTRATION of both the SOLUTIONS is same.
7.

An aqueous solution containing one gram of urea boils at 100.25^@C. The aqueous solution containing 3gm of glucose in the same volume will boil at

Answer»

`100^(@)C`
`100.25^(@)C`
`100.5^(@)C`
`100.75^(@)C`

ANSWER :B
8.

An aqueous solution containing a mixture of copper (II), iron (II) and lead (III) ions was treated with an excess of aqueous ammonia. What precipitate was left by this reaction,

Answer»

COPPER (II) HYDROXIDE only
IRON (II) hydroxide only
Lead (II) hydroxide only
Lead (II) hydroxide and iron (II) hydroxide

ANSWER :D
9.

An aqueous solution containing 6.5 g of NaCl of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is

Answer»

`2000CM^(3)`
`100CM^(3)`
`200cm^(3)`
`1000cm^(3)`

Solution :WEIGHT of `NaCl=6.5xx(90)/(0.1)=5.85g`
No. of equivalents of NaCl
`=("wt. of NaCl")/("Mol. Mass of NaCl")=(5.85)/(58.5)=0.1`
No. of equivalents of NaOH obtained=0.1
`thereforeN_(NaOH)xxV_(NaOH)=N_(CH_(3)COOH)xxV_(CH_(3)COOH)`
`implies0.1xx1000=1xxV_(CH_(3)COOH)`
`impliesV_(CH_(3)COOH)=100cm^(3)`
10.

An aqueous solution containing 6.5 g of NaCl of 90% purity was subjected to electrolysis.After the complete electrolysis, the solution was evaporated to get solid NaoH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is

Answer»

`2000 cm^(3)`
`100 cm^(3)`
`200 cm^(3)`
`1000 cm^(3)`

SOLUTION :WEIGHT of NaCl`=6.5xx90/100=5.85g`
No. of equivalents of NaCl
`=("WT. of " Nacl)/("mol. Mass of" NaCl)=5.85/58.5=0.1`
No. of equivalents of NaOH OBTAINED =0.1
`:.N_(NaOH)xxV_(NaOH)xxN_(CH_(3)COOH)xxV_(CH_(3)COOH)`
`implies0.1xx1000=1xxV_(CH_(3)COOH)`
`impliesV_(CH_(3)COOH)=100cm^(3)`
11.

An aqueous solutioncontaining 5% by weight of urea and 10% by weight of glucose . What will be the DeltaT_(f) of the solution (K_(f) for H_(2)O is 1.86^(0)K-molal)

Answer»


ANSWER :C
12.

An aqueous solution containing 288 g of a nonvolatile compound having composition C_nH_(2n)O_n in 90 g of water boils at 101.24^@C at 1-atm pressure. What is the molecular formula of the compound? K_b = 0.512^@ C/m.

Answer»

SOLUTION :`C_44H_88O_44 `
13.

An aqueous solution containing 1g of urea boils at 100.25^(@)C. The aqueous solution containing 3g of glucose in the same volume will boil be

Answer»

`100.75^(@)C`
`100.5^(@)C`
`100.25^(@)C`
`100^(@)C`

ANSWER :C
14.

An aqueoussolution containing 1 mol of HgI_(2) and 2 molof Nal is orange in colour .Onadditionod excessNaI , thesolutionbecomescolourless .Theorangecolour reappearsonsubsequent addition of NaOCl.

Answer»

Solution :`HgI_(2)` is having orange / scarletcolour it dissolvein sodiumiodidedue to theformation of a complex .
`HgI_(2) + 2Nal RARR Na_(2)HgI_(4)`
with `NaOCI` AGENT a PRECIPITATE of `HgI_(2)` is FORMED
`3Na_(2)HgI_(4) + 2NaOCI + 2H_(2)O rarr 3HgI_(2) + 2NaI_(3) + 4NaOH + 2NaCI`
15.

An aqueous solution containing 12.48 g of barium chloride in 1.0 kg of water boils at 373.08321K.Calculate the d.egree of dissociation of barium I chloride. (Given K_(b), of H_(2)O = 0.52 K m^(-1), molar mass of Ba CI_(2) = 208.34 g mol^(-1)]

Answer»

Solution :`W_(R) = 12.48 ` G of `BaCI_(2)`
`W_(A) = 1 `kg
Boiling POINT of solution = 373.0832 K
Degree of DISSOCIATION of `BaCI_(2)`= ?
`K_(b)=0.52 km^(-1)`
`M_(B)= 208.34` g/mol .
Boiling point of water =373 K
`T_(b) = iK_(b)m`
`0.0832 = ixx0.52xx(12.48)/(208.34) xx1 kg `
`i= (208.34xx0.0832)/(0.52xx12.48)`
`=(17.33)/(6.48)`
i = 2.67
`BaCI_(2)= Ba^(2+)+ 2CI^(-) `
1 mol `""` 0 `""` 0
`1-alpha "" alpha "" 2 alpha`
Total 1 + 2 `alpha`
`i =1 +2 alpha ` or `alpha = (i-1)/(2) = (2.67-1)/(2)`
`= (1.67)/(2) = 0.835`
16.

An aqueous solution containing 0.25 moles of a strong electrolyte A in 500 g of water freezes at -2.8^(@)C. How many ions are formed per formula unit of A (K_(f)=1.86^(@)C) ?

Answer»

1
2
3
4

Solution :`DeltaT_(F)=2.8^(@)C`
`DeltaT_(f)` (OBSERVED) `= iK_(f) XX m`
`2.8= ixx 1.86 xx 0.50`
or `i=(2.8)/(1.86 xx 0.50)=3`
17.

An aqueous solution containing 0.10 g KCIO_3(formula weight = 214.0) was treated with an excess of KI solution. The solution was acidified with HCl. The liberated I_2consumed 45.0 mL of thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution.

Answer»

SOLUTION :`KIO_3 + 5KI + 3K_2O + 3I_2`
0.062 M
18.

An aqueous solutio of a salt (A) gives white precipitate with NaCl solution and filtrate of this with H_(2)S gives black precipitate . Then which one of the following cations should be present in

Answer»

`HG^(+)`
`CU^(+)`
`Cu^(++)`
`PB^(++)`

ANSWER :D
19.

An aqueous solutin containing compounds A and B shows optical activity A and B are stereoisomers. Which of the following possibilities cannot be correct ?

Answer»

A has TWO chiral centers, but B does not have any because it has a symmetry plane
A and B are enantiomers
A and B are diasteromers
A and B are not present in equal amounts

Solution :COMOPOUND 'B' MAY be MSO but MESO has plane of symmetry with chiral centres.
20.

An aqueous sol^(n) containg 10g ofmixture of urea and glucose boils at 100.58^(@)C. Addition of a further 6.0g glucose to the above solution causes itto boilat 100. 77^(@)C mass percentage of urea in the origin mixtureis 7y then y=?

Answer»


ANSWER :2
21.

An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is :

Answer»

`40 mL`
`20 mL`
`10 mL`
`4 mL`

Solution :Equivalent mass of OXALIC acid
`=126/2=63`
Normally of exalic acid
`=6.3/63xx1000/250=0.4N`
`underset(("oxalic acid"))(N_(1)V_(1))=underset((NaOH))(N_(2)V_(2))`
`0.4xx10=0.1xxV_(2)`
`V_(2)=(0.4xx10)/(0.1)=40ML`
22.

An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because …….. .

Answer»

`[Co(H_(2)O)_(6)]^(2+)` is transformed into `[CoCl_(6)]^(4-)`
`[Co(H_(2)O)_(6)]^(2+)` is transformed into `[CoCl_(4)]^(2-)`
TETRAHEDRAL complexes have smaller crystal field splitting than octahedral complexes
tetrahedral complexes have larger crystal field splitting than octahedral COMPLEX.

ANSWER :B::C
23.

An aqueous of salt (A) gives a whitecrystallineprecipitate (B) with NaCI solution.The fitrategives a blackprecipitate (C) when H_(2)S is passedthrough it compound (B) dissolve in hot waterand thesolution givesyellowprecipitate (D) antratement with potassiumiodide and on cooling .The compound (A) doesnot give are gas with dilute HCIbut liberatesa reddishbrowngas on heatingidentifythecompounds (A) to (D) givingthe involvedequations.

Answer»


Answer :A: `PB(NO_(3))_(2)` B: `PbCI_(2)` C: `PBS` D: `PbI_(2)`
24.

An aqueous of glucose C_(6)H_(12)O_(6) has an osmotic pressure of 2.72 atmospheres at 298 K. How many moles of glucose were dissolved per litre of the solution ? (R = 0.082 lit. atm. "mol"^(-1)"deg"^(-1))

Answer»

SOLUTION :0.1113 MOL
25.

An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCI. This is because …………… .

Answer»

`[Co(H_(2)O)_(6)]^(2+)` is TRANSFORMED into `[CoCl_(6)]^(4-)`
`[Co(H_(2)O)_(6)]^(2+)` is transformed into `[CoCl_(4)]^(2-)`
tetrahedral complexes have samaller crystal field splitting than OCTAHEDRAL complexes
tetrahedral complexes have LARGER cystal field splitting than octahedral complex.

Answer :B::C
26.

An aqueous blue coloured solution of a trasition of metal sulphate reacts with H_(2)S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involvedin the formation of A and B.

Answer»
27.

An aqueous blue-coloured solution of a transition metal sulphate reacts with H_2S I acidic medium to give a black precipitate A which is insoluble in warm auqous solution of KOH. The blue solution on treatment with KI in weakly acidic medium turns yellow and produces a white precipitate B. Identify the transition metal ion. write the chemical reactions involved in the formation of A and B.

Answer»

Solution :An aqueous blue-COLOURED solution of a TRANSITION metal sulphate reacts with `H_2S` in acidic medium to give a black precipitate A which is insoluble in warm aqueous solution of KOH. It must be the black precipitate of CuS because copper sulphate is also blue coloured. The blue solution on treatment with KI in a weak acidic medium TURNS yellow iodide is first formed which decomposes to give white cuprous iodide and iodine.
`CuSO_4+H_2Soverset("Acidic medium")toCuSdarr+H_2SO_4`
`CuSO_4+2KItoCuII_2+K_2SO_4` (Yellow SOLN.)
`2CuI_2tounderset((B) "white PPT. insoluble in H_2O")(Cu_2I_2darr+I_2)`
28.

An aquaeous solutin of X is added slowly to an aqueous solutions of Y as in List I. The variatin in conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using code given below the lists: Lists I "P."underset(X)((C_(2)H_(5))_(3)N)+underset(Y)(CH_(3)COOH) "Q."underset(X)(KI(0.1M))+underset(Y)(AgNO_(3)(0.01M)) "R."underset(X)(CH_(3)COOH)+underset(Y)(KOH) "S."underset(X)(NaOH)+underset(Y)(HI) List II 1. Conductivity decreases and then increases 2. Conductivity decreases and then does not change much 3. Conductivity increases and then does not change much 4. Conductivity does not change much and then increases

Answer»

`{:("P","Q","R","S"),("2","3","1","4"):}`
`{:("P","Q","R","S"),("3","2","1","4"):}`
`{:("P","Q","R","S"),("2","3","4","1"):}`
`{:("P","Q","R","S"),("3","2","4","1"):}`

ANSWER :A
29.

An aqueous blue coloursolutionof atransitionmetal sulphate reacts with H_(2)S in acidicmedium to give a blackprecipitate A which is insolution in warm aqueoussolution of KOH .The bluesolutionon tretmentwith KJ inweakly acidic mediumturns yellowand producesa whiteprecipitate B identify the transitionmetalion write thechemicalreactioninvolved in theformation of A and B

Answer»

Solution :An aqueousblue -colouredsolutionof atrastionmetal sulphatereacts with `H_(2)S` in acids medium to GIVE a black precipitate A which is insolublein warmaqueoussolution of `KOH` it most be ablackprecipitate of CuS becausecoppersalphate is bluecolouredalso .Theblue SOLUTIONON troatmentwith KI inweaklyacidmediumfurmyellowand producess a WHITEPRECIPITATE B inthisreaction copric iodide is formedfirst whichdecomposess to givewhitecuposeiodide iodide
`CuSO_(4) + H_(2)S OVERSET("Acidic medium")rarr underset(Black)(CuSdarr) +H_(2)SO_(4)`
`CuSO_(4) +2KI rarr CuI_(2) +K_(2)SO_(4)`
`2CuI_(2) rarr Cu_(2)I_(2) darr +I_(2)`
30.

An aqueolis solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is

Answer»

40 mL
20 mL
10 mL
4 mL

Answer :A
31.

An aqqeoussolution of a substance gives a white precipitate on tretment with diute hydrocloric acid which which dissolves on heating .When hydrogen sulphide ispassed through the hot acidic solution a black precipitateis obtained .The substanceis a

Answer»

`Hg_(2)^(2+)` SALT
`Cr^(+)` salt
`AG^(o+)` salt
`Pb^(2+)`salt

Solution :A whiteprecipitate which issolublein hot WATER is LEAD chloride
`Pb^(2+) + 2HCI rarrunderset(white)(PbCI_(2) darr ) + 2H^(oplus)`
`underset(Hot solution)(PbCI_(2))+H_(2)S overset(H^(o+))rarr overset(White)underset(Black)(PbSdarr) +2HCI`
32.

An aq. solution of an inorganic compound (X) shows the following reactions. It liberates I_(2) from acidified KI solution.

Answer»

Solution :`X-H_(2)O_(2)`
`H_(2)O_(2)+2KI+H_(2)SO_(4) rarr I_(2) UARR +K_(2)SO_(4)+2H_(2)O`
33.

An aq. solution of an inorganic compound (X) shows the following reactions. It is used to restore old oil paintings. Identify (X) and give chemical reactions for the steps (i) to (iv).

Answer»

Solution :`X-H_(2)O_(2)`
`4H_(2)O_(2)+PBS RARR underset("white")(PbSO_(4))+4H_(2)O`
34.

An aqaeoussolutionof FeSO_(4).Al_(2)(SO_(4))_(2) and chrome alum is heated with excess of Na_(2)O_(3) and chrome alumin heatedwith excess of Na_(2)O_(2) and fiterad .The matrialsobtained are

Answer»

A colourless fitrateand a GREEN residue
A yellow fitrateand a green residue
A yellow fitrateand a brown residue
A green fitrateand a green brown

Solution :In the presence of peroxide, chromium IONS are oxidisedto chromate ions which GIVE a yellowfitrate ferrie ions form brownprecipitate of `FE(OH)_(3)`
35.

An aq. solution containing 64% by weight of volatile liquid 'A' (molecular mass 128) has pressure of 145mm, then vapour pressure of 'A' is (V.P. of H_2O is 155 mm)

Answer»

150 mm
145 mm
105 mm
21 mm

Answer :C
36.

An appropriate reagent for the conversion of 1-propanol to 1-propanal is

Answer»

acidified potassium DICHROMATE
ALKALINE potassium permaganate
pyrindium CHLOROCHROMATE
acidified `CrO_(3)`

Answer :C
37.

An apparatus used for the measurement of quantityof electricity is known as a

Answer»

Calorimeter
Cathetometer
Coulometer
Colorimeter

Solution :CU voltmeter or Cu or AG coulometer are used to DETECT the amount deposited on an ELECTRODE during passage of known CHARGE through solution .
38.

Anapparatus used for the measurement of quantity of electricity is known as a

Answer»

CALORIMETER
Cathetometer
Coulometer
Colorimeter

Solution :CU VOLTMETER or Cu or Ag coulometer are used to detect the amount deposited on an electrode during PASSAGE of known charge through solution.
39.

An apparatus used for the measurement of quantity of electricity is known as a :

Answer»

Calorimeter
Cathetometer
Coulometer
Colorimeter

Answer :C
40.

An antipyretic is:

Answer»

Quinine
Paracetamol
Luminal
Piperazine

Answer :B
41.

An antipyretic is ..........

Answer»

CHLORO quinine
PARACETAMOL
morphine
ranitidine

Solution :paracetamol
42.

An antipyretic drug on prolonged usage causes irreversible kidney damage. It is

Answer»

Aspirin
Paracetmol
PHENACETIN
Pencilline

Answer :C
43.

An antioxidant which is added to butter to increase its shelf life from months to years is

Answer»

SODIUM BENZOATE
Butylated hydroxy anisole
SULPHUR dioxide
Butylatedhydroxy tolunene

Answer :B
44.

An antifriction alloy made up of antimony with tin and copper, which is extensively used in machine bearings is called-

Answer»

Duralumin
Babbitt METAL
Spiegeleisen
Amalgams

ANSWER :A::B
45.

An antigen develops antibodies which protect the body from their harmful effects. The antibodies are :

Answer»

Immunoglobulins
Phospholipids
Albumins
Lymphocytes

Answer :A
46.

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_(2)H_(6)O_(2)) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^(-1), then what shall be the molarity of the solution ?

Answer»

SOLUTION :Molar mass of ethylene glycol `[C_(2)H_(4)(OH)_(2)]`
`= 2xx12+6xx1+2xx16`
= 62 g `MOL^(-1)`
Number of moles of ethylene glycol
`= (222.6g)/(62 g mol^(-1))`
= 3.59 mol
Therefore, molality of the solution
`= (3.59 mol)/(0.20 kg)`
= 17.95 m
Total mass of the solution `= (222.6+200)g`
= 422.6 g
Given,
Density of the solution `=(422.6 g)/(1.072 g mL^(-1))`
= 394.2 mL
`= 0.3942xx10^(-3)L`
MOLARITY of the solution `= (3.59 mol)/(0.39422xx10^(-3)L)`
= 9.11 M.
47.

An antifriction alloy made up of antimony with tin and copper, which is extensively used in machiine bearings is called :

Answer»

DURALUMIN
Babbittt METAL
SPIEGELEISEN
Amalgams

ANSWER :B
48.

An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C_2H_6O_2), and 200 g ofwater. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^(-1) , thenwhat shall be the molarity of the solution ?

Answer»

Solution :Given : Mass of the solute, `C_2H_6O_2 = 222.6 g`
MOLAR mass of `C_2H_6O_2 = 62 g "mol"^(-1)`
NUMBER of moles of the solute =`(222.6 g)/(62g "mol"^(-1)) = 3.59`
Mass of the solvent = 200 g = 0.200 kg
Molality of the solution = 3.59 moles/0.200 kg = 17.95 m
Total mass of the solution = 222.6 + 200 = 422.6 g
Volume of the solution = Mass of solution / DENSITY of solution
` = (422.6 g)/(1.072g mL^(-1)) = 394.2 mL = 0.3942 L`
Molarity of the solution = `(3.59"moles")/(0.3942 L) = 9.11 M`
49.

An antifreeze solution is prepared from 222.6 g of ethylene glycol [C_2H_4 (OH)_2] and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g mL^(-1) , what will be the molarity of the solution ?

Answer»

Solution : APPLYING the following relation and SUBSTITUTING the VALUES, we get
`m = (W_B)/(M_B) XX (1000)/(W_A)= (222.6)/(62) xx 1000/200 = 2226/124 = 17.95`mol/kg
Molality (m) and molarity (M), are RELATED as follows :
`m = (M)/(d - M xx M_B xx 10^(-3))`
` 17.95 = (M)/(1.072 - M xx 62 xx 10^(-3))`
`17.95 xx 1.072 - 17.95 xx 62 xx 10^(-3) M = M`
`M (1+ 17.95 xx 62 xx 10^(-3)) = 19.24 " or "M = (19.24)/(2.11) = 9.12` mol/litre
Molarity of the solution = 9.12 M.
50.

An antifreeze solution is prepared by dissolving 31 g of ethylene glycol (Molar mass = 62 "g mol"^(-1)) in 600 g of water. Calculate the freezing point of the solution. (K_(f) for water =1.86" K kg mol"^(-1))

Answer»

SOLUTION :Given : `w_(1)=600" g,"w_(2)=31" g,"M_(2)=62" g mol"^(-1)`
Molality of the solution `=(w_(2)//M_(2))/(w_(1)//1000)=(31//62)/(600//1000)=(31)/(62)xx(1000)/(600)=5/6m`
Apply the RELATION :
`DeltaT_(f)=K_(f)m`
`=1.86" K kg mol"^(-1)xx5/6" mol kg"^(-1)`
`=1.86xx5/6=1.55`
FREEZING point of the solution `=-1.55^(@)C` or `271.45K`