52671 + Interview Questions in Chemistry in Class 11

1.	The equilibrium constant of a reaction is 73. Calculate standard free energy change.
Answer» Both A & R are true and R is the correct EXPLANATION of A Both A & R are true but R is not the correct explanation of A A is true but R is FALSE A is false but R is true ANSWER :D

Discussion

2.	(a) Stability of a crystal is reflected in the magnitude of its melting point'. Comment (b)The melting point of some compounds are given below: Water = 273 K, Ethyl alcohol =155.7 K, Diethyl ether =156.8 K, Methane =90.5 K. What can you say about the intermolecular forces between these molecules ?
Answer» SOLUTION :(a) Higher the MELTING point, greater are the forces holding the constituent particles together and hence greater is the stability (b) The intermolecular forces in water and ethyl alcohol are mainly the hydrogen bonding Higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not a strong as in water molecules. DIETHYL ether is a polar molecule. The intermolecular forces present in them are dipole dipole attraction Methane is a non-polar molecule. The only forces present in them are the weak van DER Waals forces (London dispersion forces)

Discussion

3.	(a) ' Stability of a crystal is reflected in the magnitude of its melting point . Comment (b) The melting points of some compounds are given below : water = 273 K ,Ethyl alcohal= 155.7 K , Diethyl ether = 156.8 K , Methane = 90.5kK what can you say about the intermolecular forces between these molescular ?
Answer» Solution :(a) Higher the METING point, greater are the forces holding the consituent particles together and hence greater is the stability. (b)The intermolecular forces in water and entyl alcohalare mainly the hydrogen BONDING. Higher MELTING point of water than alcoholshows that hydrogen bonding in ETHYL alcohol molecules is not are strong as in water molecules. Diethyl ether is a POLER molecule. The intermolecular forces present in them are dipolediple attraction. Methane is a non-polar molecular.The only forces present in them are the weak van der Waal's forces ( London dispersion forces).

Discussion

4.	A square planar complex is formed by hybridisation of which of the following atomic orbitals ?
Answer» <P>`s,p_x,p_y ,d_(YZ)` `s,p_x ,P_y ,d_(x^2 -y^2)` `s,p,p_y,d_(z^2)` `s,p_y ,P_z ,d_(XY)` ANSWER :B

Discussion

5.	A spinal is an important class of oxides consisting of two types of metal ions with oxides ions arranged in CCP layer. The normal spinal has one-eighth of the tetrahedral holes occupied by one type of metal ion and one-half of the octahedral holes occupied by another type of metal ion. such a spinal is formed by Zn^(2+), Al^(3+) and O^(2-) in the tetrahedral holes. if formula of the compound is Zn_(x)Al_(y) O_(z), then find the value of (x + y + z) ? The type of packing generated by type (I) is:
Answer» HEXAGONAL close packing squre close packing cubic close packing body centered packing Solution :`O_(4) Zn_((1)/(8) xx 8) Al_((1)/(2) xx 4)` So, FORMULA is `ZnAl_(2)O_(4)`

Discussion

6.	For a spontaneous process, in a reaction
Answer» which NEEDS some initaiation LIKE HEAT or energy which takes place instantaneously which takes place by itself takes place by itself or by INITIATION. Answer :D

Discussion

7.	A spherical balloon of 21 cm diameter is to be filled with hydrogen at NTP from a cylinder containing the gas at20 atmosphere at 27^(@)C.If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up.
Answer» Solution :Volume of the balloon `=(4)/(3)pi r^(3)=(4)/(3)XX(22)/(7)xx((21)/(2))^(3)=4851 cm^(3)` Volume of the cylinder =2.82 litres=2820 `cm^(3)` Pressure =20 ATM. Temperature =300 K Converting this to the volume at NTP, `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)):. (20xx2820)/(300)=(1xxV_(2))/(273)"or"V_(2)=51324" cm"^(3)`. When the pressure in the cylinder is REDUCED to one atmosphere,no more `H_(2)` will be released and hence 2820 `cm^(3)` of `H_(2)` will be left in it. Hence, volume of `H_(2)`used in filling the balloons `=51324-2820 cm^(3)=48504 cm^(3)`. Number of balloons filled =48504/4851=10

Discussion

8.	A spectral line of hydrogen with lambda = 4938 Å belongs to the series
Answer» Lyman Balmer Paschen Pfund Answer :B

Discussion

9.	A specisthat. Donatesthe electronpair pf H^(@)is termedas a / an_______.
Answer» NUCLEOPHILE BASE electrophile acid Answer :B

Discussion

10.	A specific cloud is formed on antarctia in the winter is called .........
Answer» SOLUTION :NACREOUS CLOUDS

Discussion

11.	A species ‘X’ contains 20 protons and 18 electrons. (A) species ‘Y’ contains 18 protons and 18 electrons. What are ‘X’ and ‘Y’ respectively ?
Answer» CA and Ar `Ca^(2+)` and `S^(2-)` `Ca^(2+)` and `CI^(-)` `Ca^(2+)` and Ar Solution :(i) X has 20 protons `therefore` ATOMIC number = Z = 20`therefore`The atom is Ca But it has 18 electrons `therefore` X MUST be `Ca^(2+)` (II) Y has 18 protons `therefore`atomic number = 18 = Z `therefore` The atom is Ar But no. of electron in it are 18 Y must be A

Discussion

12.	A spark plug is not necessary in a diesel engine because
Answer» DIESEL is more VOLATILE than petrol diesel has a lower ignition temperature than petrol calorific VALUE of diesel is more than that of petrol None of these Solution :DUE to higher compression ratio in diesel ENGINE the compressed air gets heated up to such a high temperature that the sprayed fuel catches fire on its own

Discussion

13.	A sparingly soluble salt having general formula A_(x)^(p+) B_(y)^(q-) and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.
Answer» <P> Solution :Suppose molar solubility of `A_(X)^(p+) B_(y)^(Q-)` is S mol `L^(-1)`. Then `{:(A_(x)^(p+)B_(y)^(q-) ,hArr,x A^(p+),+ ,yB^(q-),,),(,,x S,,y S,,):}` `K_(sp) = [ A^(p+)]^(x) [B^(q-)]^(y)=[x S]^(x) [ y S ] ^(y)= x^(x) y^(y) S^(x+y)`

Discussion

14.	A sparingly soluble salt having general formula A_x^(p+) B_y^(q-)and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between solubility and the solubility product for such salt.
Answer» Solution :A sparingly soluble salt having general formula `A_x^(p+) B_y^(q-)` . Its MOLAR solubilityis S mol `L^(-1)` . Then, `A_x^(p+) B_y^(q-)hArr xA_x^(p+)(AQ)+yB_y^(q-)` (aq) S moles of `A_xB_y` dissolve to give X moles of `A^(p+)` and y moles of `B^(q-)` Therefore , solubility PRODUCT `(K_(sp))=[A^(p+)]^x [B^(q-)]^y` `=[xS]^x [yS]^y` `=x^x y^y S^(x+y)`

Discussion

15.	A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Q_(sp)) becomes greater than its solubility product. If the solubility of BaSO_(4) in water is 8xx10^(-4)mol dm^(-3) , calculate its solubility in 0.01 mol dm^(-3) of H_(2)SO_(4).
Answer» Solution :Solubility of `BaSO_(4) ` in water `(S) = 8 xx 10^(-4) ` MOL `dm^(-3)` LTBGT `BaSO_(4) hArr Ba^(2) + SO_(4)^(2-)` `:. K_(sp) ` for `BaSO_(4) = [Ba^(2+)][SO_(4)^(2-)]=(8xx10^(-4))(8xx10^(-4))=64xx10^(-8)` As `H_(2)SO_(4)` IONIZES completely as `H_(2)SO_(4) rarr 2H^(+) + SO_(4)^(2-),[SO_(4)^(2-)]` producedfrom 0.01 M `H_(2)SO_(2-)]` produced from 0.01 M `H_(2)SO_(4)=0.01 M `. Thus, if S is the solubility of `BaSO_(4) ` in `H_(2)SO_(4) ` , then `K_(ap) = [Ba^(2+)][SO_(4)^(2-)] = S (S+ 0.01 ) = 64 xx 10^(-8) `(calculated above ) or `S^(2)+0.01 S - 64 xx 10^(-8) = 0 ` `:. S= (-0.01 PM sqrt((0.01)^(2)+4xx64xx10^(-8)))/(2) = (-0.01 pm sqrt(10^(-4)+256xx10^(-8)))/(2)` `=(-0.01 pm sqrt(10^(-4)(1+256xx10^(-2))))/(2) = (-0.01 pm 10^(-2) sqrt(1+256xx10^(-2)))/(2)` `=(-0.01 pm 10^(-2)sqrt(1.256))/(2) = (-10^(-2)+1.12xx10^(-2))/(2)` `=((-1+1.12)xx10^(-2))/(2)` `=((-1+1.12)xx10^(-2))/(2) = (0.12)/(2) xx 10^(-2) = 6 xx 10^(-4) ` mol `dm^(-3)`.

Discussion

16.	A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution Q_((sp)) becomes greater than its solubility product. If the solubility of BaSO_4 in water is 8 xx 10^(-4) "mol dm"^(-3). Calculate its solubility in 0.01 mol "dm"^(-3) of H_2SO_4.
Answer» Solution :`BaSO_(4(g)) hArr Ba_((aq))^(2+) + SO_(4(aq))^(2-)` `K_(sp)` for `BaSO_4 = [Ba^(2+)] [ SO_4^(2-) ] =sxxs =S^2` But `S=8XX10^(-4) "mol dm"^(-3)` `THEREFORE K_(sp)=(8xx10^(-4))^2 = 64xx10^(-8)` In the presence of 0.01 `MH_2 SO_4`, the expression for `K_(sp)` will be `K_(sp)=[Ba^(2+)][SO_4^(2-)]` `K_(sp)`=(s)(s+0.01)(0.01M `SO_4^(2-)` IONS from 0.01 M `H_2SO_4` ) 64 x 104 = s. (s+0.01) `s^2=0.01s-64xx10^(-8)=0` `s=(-0.01pmsqrt((0.01)^2+(4xx64xx10^(-8))))/2` `=(-0.01pm sqrt(10^(-4)+256xx10^(-8)))/2` `=(-0.01 PM sqrt(10^(-4)(1+256 xx10^(-4))))/2` `=(-0.01 pm 10^(-2) sqrt(1+0.0256))/2` `=(10^(-2) (-1pm1.012719))/2` `=5xx10^(-3) (-1+1.012719)` `=6.4xx10^(-5) "mol dm"^(-3)`

Discussion

17.	Asolution with pH 2.699 is diluted two times, then calculate pH of the resulting solution. [Given antilog of 0.3010 = 2
Answer» Solution :NEW pH = 2.699 + `log (2V)/( V) =3`

Discussion

18.	A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)). For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-), with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y). While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp), value at a particular temperature, then precipitation occurs. The solubility of BaSO_(4), in 0.1 M BaCl_(2), solution is (K_(sp), of BaSO_(4), = 1.5 xx 10^(-9))
Answer» ` 1.5 xx 10^(-9) M` `1.5 xx10^(-8) M` ` 2.25 xx 10^(-16)M` ` 2.25 xx 10^(-18) M` Solution :` BaSO_4 hArr Ba^(+2)+SO_4^(-2) ` ` 0.1 M BaCl_2 =0.1 M Ba^(+2) ` ` [SO_4^(-2)]= S, [Ba^(+2) ]= (0.1 +S) ~~ 0.1M` ` K_(sp)=[Ba^(+2) ] [SO_4^(-2) ], 1.5 xx 10 ^(-9) =(0.1 ) (S) ` `RARR S = 1.5 xx 1 0^(-8) `

Discussion

19.	A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)). For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-), with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y). While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp), value at a particular temperature, then precipitation occurs. The solubility of PbSO_(4), in water is 0.303 g/l at 25^(@)C, its solubility product at that temperature is
Answer» ` 10^(-4)M^(2) ` ` 9.18 xx 10^(-4) M` ` 10^(-6)M^(2)` ` 9.18 xx 10^(-8) M^(2)` Solution :` S= 0.303 g // lit ` ` MOL . WT = 207 + 32 + 64 =303 ` MOLES ` =( 0.303)/( 303)=10 ^(-3) ` ` S = 10 ^(-3)" MOLE " // "lit"` ` pbSO_4 hArr Punderset(S) b^(+2)+Sunderset( S) O ^(-2), S =10 ^(-3) ` ` K_(sp)= ( 10 ^(-3) M)^(2) =10 ^(-6)M^(2) `

Discussion

20.	A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+) is treated with 10^(-16)M sulphide ion. If K_(sp) od MnS, ZnS and HgS are 10^(-15),10^(-25),10^(-20)and10^(-54) respectively, which one will precipitate first ?
Answer» FeS MnS HgS ZnS Solution :Ionic product in the solution `=10^(-3)xx10^(16)=10^(-19)` The metal sulphide having the lowest solubility will precipitate FIRST provided the ionic product is higher than the `K_(SP)`. Here, all salts are of the same valence-type. So the sulphide having the lowest `K_(sp)` value will precipitate first provided `K_(sp)lt10^(-19)`, HgS has lowest `K_(sp)` value `(10^(-54))`, so will precipitate first.

Discussion

21.	A solution which is 10^(-3) M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20) and 10^(-54) respectively, which one will precipitate first ?
Answer» FeS MgS HgS ZnS Solution :`HgS` will be PRECIPITATED FIRST because [its `K_(sp)`] is MINIMUM.

Discussion

22.	A solution of which substance can best be used as both titrant and its own indicator in an oxidation-reduction titration ?
Answer» `I_(2)` NaOCl `K_(2)Cr_(2)O_(7)` `KMnO_(4)` SOLUTION :N//A

Discussion

23.	A solution of weak acid was titrated with base NaOH. The equivalence point was reached when 36.12 mL of 0.1M NaOH have been added. Now 18.06mL 0.1M HCI were added to titrated solution, the pH was found to be 4.92. What is K_(a) of acid ?
Answer» ANSWER :`1.2xx10^(-5);`

Discussion

24.	A solution of volume V contains a mole of MCI and b mole of NCI where MOH and NOH are two weak bases having dissociation constants K_(1) and K_(2) respectively. Show that the pH of the solution can be expressed as pH=(1)/(2)log_(10) [(K_(1)K_(2))/(K_(w))xx(V)/(aK_(2)+bK_(1))]
Answer»

Discussion

25.	A solution of specific gravity 1.6 g mL^(-1) is 67% by weight. What will be the % by weight of the solution of same acid if it is diluted to specific gravity 1.2 g mL^(-1)?
Answer» ANSWER :`29.78%` ;

Discussion

26.	A solution of sodium salt of unknown anion when treated with Magnesium chloride solution gives white precipitate only on boiling. The anion is
Answer» `SO_(4)^(2-)` `HCO_(3)^(-)` `CO_(3)^(2-)` `NO_(3)^(-)`. Solution :BICARBONATES on heating form insoluble carbonates `MgCl_(2) + 2NaHCO_(3)rarr2NaCl+Mg (HCO_(3))_(2)` `Mg(HCO_(3))_(2)overset(Delta)rarrMgCO_(3)darr+CO_(2)+H_(2)O`

Discussion

27.	A solution of sodium metal is liquid ammonia is strongly reducing due to the presence of
Answer» SODIUM atoms Sodium hydride Sodium amide SOLVATED ELECTRONS. SOLUTION :A solution of sodium metal in ammonia CONTAINS solvated electrons.

Discussion

28.	A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of....
Answer» SODIUM ATOMS sodium hydride sodium AMIDE SOLVATED electrons Answer :B

Discussion

29.	A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of
Answer» SODIUM atoms sodium hydride sodium amide solvated electrons Answer :D

Discussion

30.	Why does the solution of sodium in liquid ammonia possess strong reducing nature?
Answer» SODIUM atoms sodium HYDRIDE sodium amide SOLVATED ELECTRONS. Solution :solvated electrons.

Discussion

31.	A solution of silver nitrate was stirred with iron rod will it cause any change in the concentration of silver and nitrate ions ?
Answer» Solution :Since `E^(@)` of `Fe^(2+)//Fe` (-0.44 V) is LOWER than that of `Ag^(+)//Ag(+0.80 V)` ELECTRODE therefore `Ag^(+)` gets REDUCED and Fe gets oxidised As a result concentration of `Ag^(+)` ions DECREASES while that of `NO_(3)^(-)` ions remains unchanged `2 Ag^(+) (aq)+Fe(s)rarr 2 Ag(s)+Fe^(2+)(aq)`

Discussion

32.	A solution of potassium hydroxide is used to absorb carbon evolved during the estimation of carbon in an organic compound. Explain.
Answer» Solution :Carbon dioxide REACTS with `KOH` present in the solution to form SOLUBLE POTASSIUM carbonate and can be ESTIMATED. ltbr. `2KOH+CO_(2) rarr K_(2)CO_(3) +H_(2)O`

Discussion

33.	A solution of palmitic acid in benzine contains 4.24 g of acid per litre.When this solution is dropped on water surface, benzene gets evaporated and palmitic acids forms a unimolecular film on surface. If we wish to cover an area of 500 cm^(2) with unimolecular film, what volume of solution should be used? The area covered by one palmitic acid molecule may be taken as 0.21 nm^(2). Mol.wt.of palmitic acid is 256.
Answer» ANSWER :`2.386xx10^(-5)"LITRE"`;

Discussion

34.	A solution of Na_(2)S_(2)O_(3) is standardized iodometrically against 0.1262 g of KBrO_(3). This process required 45 mL of the Na_(2)S_(2)O_(3) solution. What is the strength of the Na_(2)S_(2)O_(3)? (K = 39, Br = 80)
Answer» `0.2M` `0.1M` `0.05M` `0.1N` Solution :M.eq. `Na_(2)S_(2)O_(3)` = M.eq `I_(2)` `45xxMxx1=nxx2impliesn=(45M)/(2)...(1)` M.eq `I_(2)` = M.eq `KBrO_(3)` `nxx2=6xx(0.1262)/(167)xx1000` `impliesn=(3xx0.1262xx1000)/(167)...(2)` From (1) & (2), `(45M)/(2)=(126.2xx3)/(167)impliesM=0.1`

Discussion

35.	A solution of Na_2S_2O_3 is standardised iodometrically by using K_2Cr_2O_7. The equivalent weight of K_2Cr_2O_7 in this method is:
Answer» Mol. Wt/2 Mol. Wt/6 Mol. Wt/3 equal to mol wt. Solution :`Cr_(2)O_(7)^(2-) to 2Cr^(3+)` EQ. wt. `=("Mol wt")/6`

Discussion

36.	A solution of m-chloroaniline, m-chlorophenol, m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO_(3) to give fraction A, the leftover organic phase was extracted with dil. NaOH to give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C contains respectively.
Answer» m-chlorobenzoic ACID, m-chlorophenol and m-chloroaniline m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol Solution :m-chlorobenzoic acid being the most ACIDIC can be separated by a weak base like `NaHCO_(3)` and HENCE will be labelled FRACTION A. m-chlorophenol is not as acidic as m-chlorobenzoic acid, and can be separated by a stronger base like NaOH, and hence can be labelled as fraction B. m-chloroaniline being a base, does not react with either of the bases and hence would be labelled as fraction C

Discussion

37.	A solution of HCI is prepared by dissolving 7.30 g of hydrogen chloride gas in 100 ml of water. Find the molarity of the solution. If 50.0 cm^3 of this solution is treated with 3.50 g of zinc, what volume of H_2 measured at S.T.P. will be evolved ?
Answer» ANSWER :1.12 L

Discussion

38.	A solution of H_(2)O_(2) is titrated against a solution of KMnO_(4). The reaction is : 2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) to 2Mn^(2+)+5O_(2)+8H_(2)O If it requires 46.9 mL of 0.145 M KMnO_(4) to oxidise 20 g of H_(2)O_(2), the mass percentage of H_(2)O_(2) in this solution is :
Answer» 2.9 29 21 4.9 Solution :Number of moles of `KMnO_(4)=(MV)/(1000)=(0.145xx46.9)/(1000)` `=6.8xx10^(-3)` Number of moles of `H_(2)O_(2)=6.8xx10^(-3)xx2.5=0.017` Mass of `H_(2)O_(2)=0.017xx34=0.578` Mass % of `H_(2)O_(2)=(0.578)/(20)xx100=2.9`

Discussion

39.	A solution of glycine hydrochloride contains the chloride ion and the glycinium ion, .^(+)NH_(3) - CH_(2) - COOH, which is a diprotic acid, H_(3)N^(+) -CH_(2) - COOH + H_(2)O hArr H_(3)N^(+) - CH_(2) - COO^(-) + H_(3)O^(+), K_(1) = 4.47 x10^(-3) H_(3)N^(+) - CN_(2) - COO^(-) + H_(2)O hArr H_(2)N - CH_(2) - COO^(-) + H_(3)O^(+), K_(2) = 1.66x10^(-10) Calculate pH of a 0.05 M of glycine hydrochloride.
Answer» `1.89` `2.52` `8.91` `9.18` SOLUTION :`(PH =pKa_(1) + pKa_(2))/(2)`

Discussion

40.	A solution of ferrous oxalate has been prepared by dissoliving 3.6 g L^(-1) calculate the volume of 0.01 M KMnO_(4) solution required for complete oxidatin of 100 mL of lferrous oxalte soluton in acidic medium
Answer» Solution :Molarity of `FeC_(2)O_(4)` solution `=("mass in" gL^(-1))/("MOL mass in" G mol^(-1))=(3.6)/(144)=0.025M` The balanced chemical equation for the redox reaction is : `5FeC_(2)O_(4)+3MinO_(4)^(2-)+24H^(+)rarr5Fe^(3+)+3MN^(2+)10CO_(2)+12H_(2)O` Applying molarity equation we have `(0.01xxV)/(3)(MnO_(4)^(-))=(0.025xx100)/(5)(FeC_(2)O_(4)) "in" V=(0.025xx100xx3)/(5xx0.01)=150 mL`

Discussion

41.	A solution of ethanol in water is 1.6 molal. How many grams of ethanol are present in 500 g of the solution ?
Answer» Solution :1.6 m solution would contain 1.6 MOLES of ETHANOL dissolved in 1000 g of water. Molecular mass of ethanol `(C_(2)H_(5)OH) = 46` Mass of ethanol dissolved = `46 XX 1.6 = 73.6` g Total mass of solution = `73.6 + 1000 = 1073.6` g THUS, 1073.6 g of solution contain 73.6 g of ethanol. Hence, the mass of ethanol present in 500 g of the solution `=73.6/(1073.6)xx 500 = 34.28 g`

Discussion

42.	A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is (are)
Answer» `NH_(4)NO_(3)` `NH_(4)NO_(2)` `NH_(4)CL` `(NH_(4))_(2)SO_(4)` Solution :The colourless salt H may be either `NH_(4)NO_(3)orNH_(4)NO_(2)` and the non-inflammable gas is `NH_(3)` (a) `NH_(4)NO_(3)+NaOHoverset(Delta)toNH_(3)+NaNO_(3)+H_(2)O` Non-inflammable gas `4Zn+7NaOH+NaNO_(3)overset(Delta)to4Na_(2)ZnO_(2)+NH_(3)+2H_(2)O` Sod. zincate Gas EVOLUTION restarts (b) `NH_(4)NO_(2)+NaOHoverset(Delta)toNH_(3)+NaNO_(2)+H_(2)O` `3Zn+5NaOH+NaNO_(2)overset(Delta)to3Na_(2)ZnO_(2)+NH_(3)+H_(2)O`

Discussion

43.	A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after some time. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is (are):
Answer» `NH_(4)NO_(3)` `NH_(4)NO_(2)` `NH_(4)CI` `(NH_(4))_(2)SO_(4)` SOLUTION :`NH_(4)^(+)` salts liberate `NH_(3)` with `NaOH` `NH_(4)NO_(3)` liberates `N_(2)O` (FLAMMABLE gas)

Discussion

44.	A solution of CoCI_(2).6H_(2)O in isopropyl alcohol and water is purple. The color change to blue when we add
Answer» concentrated HCI `AgNO_(3)(aq.) both (1) and (2) none of these Solution :The purple color is due to the MIXTURE of `[Co(H_(2)O)_(6)]^(2+)` (pink) and `[CoCl_(4)]^(2-)` (blue): `[Co(OH_(2))_(6)]^(2+)+4Cl^(-)hArr[CoCl_(4)]^(2-)+6H_(2)O` When we add concentrated HCI, excess `Cl^(-)` shifts the equilibrium to the right (blue). ADDING `AgNO_(3)` (aq.) removes some `Cl^(-)` by PRECIPITATION of `AgCl(s)` and favours the reaction to the left (PRODUCES more `[Co(OH_(2))_(6)]^(2+)` and the resulting solution is pink.

Discussion

45.	A solution of a non-volatile solute in water has a boiling point of 375.3K. Calculate the vapour pressure of water above this solution at 338K. Given, p_(0) (water) = 0.2467 atm at 338K and K_(b) for water = 0.52.
Answer» Solution :`DeltaT_(b)=(375.3-373.16)=2.15K` We know that, `DeltaT_(b)`= Molality `XX K_(b)` 2.15 = Molality `xx 0.52` Molality `= (2.15)/(0.52) = 4.135` i.e., 4.135 moles of the SOLUTE present in 1000g of water (55.5 moles) Mole fraction of water `= (55.5)/(4.135+55.5)=(55.5)/(59.635)` VAPOUR pressure of water above solution = Mole fraction `xx p_(0)` `=(55.5)/(59.635)xx0.2467=0.23`atm.

Discussion

46.	A solution of a metal ion when treated with KI solution gives a red precipitate which dissolves in excess of KI solution to give a colourless solution. Moreover, the solution of the same metal ion on treatment with the solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is
Answer» `PB^(2+)` `Hg^(2+)` `Cu^(2+)` `CO^(2+)` SOLUTION :It is the correct answer : `{:("Hg"^(2+),+,2l^(-),RARR,Hgl_(2),),(,,(Kl),,("Scarlet red ppt"),),(Hgl_(2),+,2Kl,rarr,underset(("Colourless"))(K_(2)Hgl_(4)),),("Hg"^(2+),+,Co(SCN)_(2),rarr,underset(("Deep blue crystalline ppt."))(Co [Hg (SCN)_(2)]),):}`

Discussion

47.	A solution of a compound X in dilute HCl on treatment with a solution of BaCl_(2) gives a white precipitate of compound Y which is insoluble in conc. HNO_(3) and and conc. HCl. Compound X imparts golden yellow colour to the flame. underset(("immparts golden yellow colour"))(X("Solution in dilute HCl"))+BaCl_(2) to underset("White")(Y)underset("Conc. HCl")overset("Conc. "HNO_(3))toInsoluble What are compounds X and Y?
Answer» X is `MgCl_(2)` and Y is `BaSO_(4)` X is `CaCl_(2) and Y` is `BaSO_(4)` X is `Na_(2)SO_(4)` and Y is `BaSO_(4)` X is `MgSO_(4)` and Y is `BaSO_(4)` Solution :`underset((X))(Na_(2)SO_(4))+BaCl_(2) to underset((Y))(BaSO_(4))+2NaCl`

Discussion

48.	A solution of a colourles salt H on boiling with excess of NaOH produces a non-flammable gas and the evolution of gas ceases after sometime. Upon addition of Zn dust to the same solution, the evolution of gas restarts. The colourless salt (S) 'H' is (are)
Answer» `NH_(4)NO_(3)` `NH_(4)NO_(2)` `NH_(4)Cl` `(NH_(4))_(2)SO_(4)` Solution :The COLOURLESS SALTS are `NH_(4)NO_(3)` (a) and `NH_(4)NO_(2)` (B). `underset((a))(Na_(4)NO_(3)) + NaOH rarr NaNO_(3) + underset(("Colourless gass"))(NH_(3) + H_(2)O)` `NaNO_(3) + 8[H] overset((Zn - "dust"))rarr NaOH + NH_(3) + 2H_(2)O` `underset((b))(NH_(4)NO_(2)) + NaOH rarr NaNO_(2) + NH_(3) + H_(2)O` `NaNO_(2) + 6[H] overset((Zn - "dust"))rarr NaOH + NH_(3) + H_(2)O`

Discussion

49.	A solution of 10 mL of (M)/(10) FeSO_(4) was titrated with KMnO_(4) solution in acidic medium, the amount of KMnO_(4) used will be :
Answer» 10 mL of 0.5 M 10 mL of 0.1 M 10 mL of 0.02 M 5 mL of 0.1 M Solution :The INVOLVED reaction is : `2KMnO_(4)+8H_(2)SO_(4)+10FeSO_(4) to 5Fe_(2)(SO_(4))_(3)+2MnSO_(4)+K_(2)SO_(4)+8H_(2)O` `(M_(1)V_(1))/(2)=(0.1xx10)/(10)` `M_(1)V_(1)=0.2` which is possible in (c )

Discussion

50.	A solution of 1 molal concentration of a solute will have maximum boiling point elevation when the solvent is :
Answer» ETHYL alcohol acetone benzene chloroform Answer :C

Discussion

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