94420 + Interview Questions in Chemistry in Class 12 Page 572 InterviewSolution

28551.	The major product of the following reaction is-
Answer» SOLUTION :

Discussion

28552.	The mole fraction of C_(2)H_(5)OH (Molar mass = 46 ) in 5 molal aqueous ethyl elchol solution is
Answer» 0.0826 0.826 5 5/55.55 Solution :5 molal aqueous ethyl ALCOHOL sol. Means 5 moles of ethyl alcohol in 1000 g of water =1000/18 moles = 55.56 moles `x_(2)=5/(5+55.56)=5/(60.56)=0.082`

Discussion

28553.	The major product of the following reaction
Answer» ANSWER :B

Discussion

28554.	The mole fraction of ethyl alcohol in its solution with methyl alcohol is 0.80. The vapour pressure of ethyl alcohol at the temperature of the solution is 40 mm of Hg. What is its vapour pressure in solution if the solution is ideal ?
Answer» <P> Solution :`p_(C_(2)H_(5)OH)=x_(C_(2)H_(5)OH)xxp_(C_(2)H_(5)OH)^(@)=0.80xx40mm=32mm.`

Discussion

28555.	The major product of the following reaction
Answer» ANSWER :B

Discussion

28556.	The major product of the following reaction
Answer» a hemiacetal an acetal an ether an ester. Solution :Dihydropyran reacts with alcohols in the presence of anhydrous `H^(+)` to FORM acetals ` THEREFORE` opteion B is true.

Discussion

28557.	The major product of the following addition reaction is :H_(3)C-CH=CH_(2) overset(Cl_(2)//H_(2)O) to
Answer» `CH_(3)-UNDERSET(CL)underset(\|)(CH)-underset(Cl)underset(\|)(CH_(2))` `H_(3)C -underset(OH)underset(\|)(CH)-underset(Cl)underset(\|)(CH_(2))` Solution :`H_3C - CH = CH_2 overset(Cl_2// H_2O)to H_3C - underset(OH)underset(\|)CH - underset(Cl) underset(\|)CH_2`

Discussion

28558.	The mole fraction of benzene in a solution in toluene is 0.50. Calculate the weight percent of benzene in the solution.
Answer» Solution :Suppose the weight percent of benzene in the solution = x. This means that in 100 g solution, `"Mass of benzene = x g,Mass of toluene "=(100-x)g` Mol. Mass of toluene `(C_(6)H_(5)CH_(3))=92` `THEREFORE""(n_(B))/((n_(B)+n_(T)))=0.50, i.e., (x//78)/(x//78+(100-x)//92)=0.50` `(x)/(78)XX(78xx92)/(92x+78(100-x))=0.50 or 92x=46x+3900-39x or 85x=3900 or x = 45.9%`. Alternatively, `(n_(B))/(n_(B)+n_(T))=0.5, i.e., (w_(B)//78)/(w_(B)//78+w_(T)//92)=0.5 or (w_(B))/(78)=0.5or(w_(B))/(78)0.5((w_(B))/(78)+(w_(T))/(92))=(1)/(2)((w_(B))/(78)+(w_(T))/(92))` `"or"(2w_(B))/(78)=(w_(B))/(78)+(w_(T))/(92)or (w_(B))/(78)=(w_(T))/(92) or (w_(T))/(w_(B))=(92)/(78) or 1+(92)/(78)` `"or"(w_(B)+w_(T))/(w_(B))=(170)/(78)"or"(w_(B))/(w_(B)+w_(T))=(78)/(170)=0.459.` Hence, `wt%=45.9.`

Discussion

28559.	The major product of the above reaction is
Answer» SOLUTION :

Discussion

28560.	The mole fraction of a solute in a solutions is 0.1. At 298K molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm^(-3). The ratio of the molecular weights of the solute and solvent, (MW_("solute"))/(MW_("solvent")) is
Answer» Solution :`M_(B)/M_(A)=9` For DETAILS, constant Example 2.20 (TEXT PART)

Discussion

28561.	The major product of the above reaction is
Answer» ANSWER :B

Discussion

28562.	The mole fraction of a solute in a solution is 0.1 At 298K molarity of this solution is the same as its molality. The density of this solution at 298K is 2.0gcm^(-3). The ratio of the molecular weights of the solute and solvent, m_("solute")//m_("solvent") is ............
Answer» SOLUTION :`(n_(1))/(n_(1)+n_(2))=0.1` and `(n_(1)xx1000)/((n_(1)M_(1)+n_(2)M_(2))d)=(n_(1)xx1000)/(n_(2)M_(2))`.

Discussion

28563.	Themajor product of reaction is :
Answer» SOLUTION :

Discussion

28564.	The mole fraction of a solute in its solution in acetic acid is 0.2. The mass of solute (molar mass = 40) in 120g. Of acetic acid would be
Answer» 2 g 8 g 10 g 20 g Solution :MOLE fraction of acetic acid = 1 - 0.2 = 0.8 :. Solution contains `0.8 xx 60 = 48 `g. of acetic acid and `0.2 xx 40 = 8 ` g of solute :. 120 g acetic acid will contain` (8 xx 120)/48=20 g.` of solute.

Discussion

28565.	The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is same as its molality Density of the solution at 298 K is "2.0 g cm"^(-3). The ratio of the molecular weights of the solute and solvent ((MW_("solute"))/(MW_("solvent"))), is
Answer» Solution :Mole fraction of solute = 0.1 means moles of solute = 0.1, moles of solvent = 0.9 `"Mass of solute "=0.1xxM_("solute")G,` `"Mass of solvent "=0.9xxM_("solvent")g` `=(0.1M_("solute")+0.9M_("solvent"))g` Total VOLUME of solution `=("Mass")/("Density")=(0.1M_("solute")+0.9M_("solvent"))/(2xx1000)L` Molarity of solution `=(0.1xx2000)/(0.1M_("solute")+0.9M_("solvent"))"mol L"^(-1)` `=(200)/(0.1M_("solute")+0.9M_("solvent"))` `"Molality of solution"` `=(0.1xx1000)/(0.9M_("solvent"))"mol KG"^(-1)=(100)/(0.9M_("solvent"))` `"As molarity = molality"` `(200)/(0.1M_("solute")+0.9M_("solvent"))=(100)/(0.9M_("solvent"))` `"or"0.1M_("solute")+0.9M_("solvent")=2xx0.9M_("solvent")` `"or"(0.1M_("solute"))/(M_("solvent"))+0.9=1.8` `"or"(M_("solute"))/(M_("solvent"))=(0.9)/(0.1)=9`

Discussion

28566.	The major product obtained when tert -butyl bromide is heated with sodium ethoxide is
Answer» 2- Methy1 -1 propene Ehene TERT -Buty 1 methy1 ether Diethy1 ether Answer :A

Discussion

28567.	The mole fraction of a solute in bezene solvent is 0.2. Themolality of the solution will be:
Answer» 14 3.2 1.4 2 Solution :`X_(B)=n_(B)/(n_(B)+n_(A))` `0.2=n_(B)/(n_(b)+1000//78)` `1/0.2=(n_(B)+12.82)/n_(B)` `5=1+(12.82)/n_(B)n_(B)=(12.82)/4=3.2` (Mass of solvent is 1000 G)

Discussion

28568.	The major product obtained when tert-butyl bromide is heated with sodium ethoxide is
Answer» 2-Methylpropene ETHENE tert-Butylmethylether Diethyl ether Answer :A

Discussion

28569.	The mole fraction of a gas dissolved in a solvent is given by Henry's law. Constant for gas in water at 298 K is 5.55xx10^(7) Torr and the partial pressure of the gas is 200 Torr, then what is the amount of the gas dissolved in 1.0 kg of water?
Answer» `2.0xx10^(-4)mol` `2.5xx10^(-5)mol` `3.7xx10^(-6)mol` `1.2xx10^(-8)mol` SOLUTION :By Henry's law, `p_(A)=K_(H)XX x_(A)` `"or"x_(A)=(p_(A))/(K_(H))=("200Torr")/(5.55xx10^(7)"Torr")=3.6xx10^(-6)` `"But"x_(A)=(n_(A))/(n_(A)+n_(H_(2)O))~=(n_(A))/(n_(H_(2)O))=(n_(A))/(1000//18)` `therefore""n_(A)=m_(A)xx(1000)/(18)=3.6xx10^(-6)xx(1000)/(18)" mole"` `=2.0xx10^(-4)"mole."`

Discussion

28570.	The major product obtained when phenol is treated with sodium hydroxide and carbon di oxide is _______.
Answer» SALICYALDEHYDE SALICYLIC ACID benzaldehyde benzoic acid Answer :B

Discussion

28571.	The mole fraction of a solute in a solution 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm^(-3). The ratio of the molecular weight of the solute and solvent, ((MW_("solute"))/(MW_("solvent"))), is
Answer» Solution :`m=(X_(A)xx1000)/(X_(B)xxM_(A))` `m=(1000)/(9M_(A))`……(i) `M=(n_(B)xx1000xxd)/(n_(A)xxM_(A)+n_(B)xxM_(B))=(X_(B)xx1000xx d)/(X_(A)xx M_(A)+X_(B)xxM_(B))` `=(200)/(0.9M_(A)+0.1M_(B))=(2000)/(9M_(A)+M_(B))`……(ii) As m = M `(1000)/(9M_(A))=(2000)/(9M_(A)+M_(B))RARR 9M_(A)+M_(B)=18M_(A)` `therefore 9M_(A)=M_(B) therefore (M_(B))/(M_(A))=9`

Discussion

28572.	The major product obtained when, chlorobenzene is reacted with chlorine in the presence of FeCl_(3)
Answer» 1, 2-dichlorobenzene 1, 3- dichlorobenzene 1, 4- dichlorobenzene 1, 2, 4- trichlorobenzne Answer :C

Discussion

28573.	The major product obtained when aniline is treated with acetic anhydride in the the presence of pyridine is……….
Answer» <P>p BROMO acetanilide p - bromo aniline acetanilide 2,4,6 TRIBROMO aniline Answer :A

Discussion

28574.	The molarity of urea (molar mass 60 g mol^(-1)) solution by dissolving 15 g urea in 500 cm^(3) of water is
Answer» 2 mol `dm^(-3)` 0.5 mol `dm^(-3)` 0.125 mol `dm^(-3)` 0.0005 mol `dm^(-3)` Solution :Urea MOLAR MASS = 60 g/mol Molarity `=W(M xx V(dm^(3)))` `=15/(60 xx 05) = 15/(6 xx 5) = 1/2` `=0.5 mol dm^(-3)`

Discussion

28575.	The major product obtained when 2-bromobutane is heated with alcoholic KOH is :
Answer» 1-Butene 2-Butene 1-Butanol 2-butanol Solution :`CH_(3)underset(Br)underset(\|)(C)HCH_(2)CH_(3)OVERSET("alc. KOH")rarrunderset("2-Butene")(CH_(3)CH=CHCH_(3))`

Discussion

28576.	The major product obtained on treatment of CH_(3)CH_(2)CH(F)CH_(3) with CH_(3)O^(-)//CH_(3)OH is
Answer» `CH_(3)CH_(2)CH(OCH_(3))CH_(3)` `CH_(3)CH=CHCH_(3)` `CH_(3)CH_(2)CH=CH_(2)` `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)OCH_(3)` Solution :According to Saytzeff.s rule, the MAJOR PRODUCT will be that one which contains more NUMBER of substituents around the double bond.

Discussion

28577.	The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :
Answer» salicylaldehyde salicylic ACID phthalic acid BENZOIC acid Answer :B

Discussion

28578.	The major product obtained on treatement of CH_(3)CH_(2)CH(F)CH_(3)with CH_(3)O^(-)//CH_(3)OH is
Answer» `CH_(3)CH_(2)CH(OCH_(3))CH_(3)` `CH_(3)CH=CHCH_(3)` `CH_(3)CH_(2)CH=CH_(2)` `CH_(3)CH_(2)CH_(2)CH_(2)OCH_(3)` Answer :C

Discussion

28579.	The molarity of pure water (density of water=1g ml^-1)
Answer» 55.55M 50M 60M 5M Answer :A

Discussion

28580.	the major product obtained on interaction of phenol with sodium hydroxide ad carbon dioxide is
Answer» salicyaldehyde SALICYLIC ACID phthalic aciid benzoic acid Solution :Salicylic acid. (KOLBE's REACTION).

Discussion

28581.	What is the molality of pure water
Answer» 55.6 18 1 5.56 Answer :A

Discussion

28582.	The major product obtained in the photobromination of 2-methylbutane is
Answer» 1-Bromo-2-methylbutane<BR>1-Bromo-3-methylbutane 2-Bromo-3-methylbutane 2-Bromo-2-methylbutane Solution :`CH_(3)-OVERSET(CH_(3))overset(\|)(CH)-CH_(2)-CH_(3) UNDERSET(-HBr)overset(Br_(2)//hv)to` `CH_(3)-underset(Br)underset(\|)overset(CH_(3))overset(\|)(C)-CH_(2)-CH_(3)`

Discussion

28583.	The major product obtained in the following reactions is
Answer» `(+)-C_(6)H_(5)CH(Ot-Bu)CH_(2)C_(6)H_(5)` `(-)-C_(6)H_(5)CH(Ot-Bu)CH_(2)C_(6)H_(5)` `(+-)-C_(6)H_(5)CH(Ot-Bu)CH_(2)C_(6)H_(5)` `C_(6)H_(5)CH=CHC_(6)H_(5)` SOLUTION :tBuOK is very strong base and is BULKY as well. UPON HEATING, it will cause elimination of alkyl halide (`E_(2)` mechanism) to form. 1,2-diphenylethene as the product.

Discussion

28584.	The molarity of orthophosphoric acid having purity of 70% by weight and specific gravity 1.54 would be
Answer» 11 M 22 M 33 M 44 M Solution :70% by weight 70 GM `H_(3)PO_(4) rarr 100` gm solution/sample `V=W/d=100/1.54""M=(70xx1000)/(98xx100//1.54)=11M`

Discussion

28585.	The molarity of HNO_(3) in a sample which has density 1.4 g/mL and mass percentage of 63% is …….
Answer» 14 12 8 6 Solution :Where, `%(w)/(w)=63%` `rho = 1.4` g/mL `M =((%(w)/(w)xx rho xx 10))/(MM)` `M=((63xx1.4xx10))/(63)` M = 14 mol/L

Discussion

28586.	The majorproduct obtained in the photo catalysed bromination of 2-methylbutane is :
Answer» 1-bromo-2-methylbutane 1-bromo-3-methylbutane 2-bromo-2-methyl butane 2-bromo-2-methyl butane Solution :The ORDER of substitution in different alkanes is `3^(@) GT 2^(@) gt 1^(@)` Thus the bromination of 2-methy lbutane MAINLY gives 2-Bromo-2- methyl butane `CH_(3)-underset(2-"methylbutane")(CH_(2))-overset(CH_(3))overset(\|)(CH)-CH_(3) overset(Br_(2))to`

Discussion

28587.	The molarity of a solution obtained by mixing 800 ml of 0.5 M HCl with 200 ml of 1 M HCl will be
Answer» 0.8 M 0.6 M 0.4 M 0.2 M SOLUTION :`underset(("Initial"))(M_(1)V_(1))+M_(2)V_(2)=underset(("FINAL"))(M_(3)V_(3))` `V_(3)=V_(1)+V_(2)=800+200=1000` `M_(3)xx1000=0.5xx800+1xx200` `M_(3)=(600)/(1000)=0.6M`.

Discussion

28588.	The major product obtained in the following reaction is-
Answer» `(+)C_(6)H_(5)CH(O^(t)BU)CH_(2)C_(6)H_(5)` `(-)C_(6)H_(5)CH(O^(t)Bu)CH_(2)C_(6)H_(5)` `(+-)C_(6)H_(5)CH(O^(t)Bu)CH_(2)C_(6)H_(5)` `C_(6)H_(5)CH = CHC_(6)H_(5)` Solution :

Discussion

28589.	The major product obtained in the following reaction is :
Answer» ANSWER :B

Discussion

28590.	The molarity of a solution obtained by mixing 800 mL of 0.5 M HCl with 200 mL of 1 M HCI will be
Answer» 0.8 M 0.6 M 0.4 M 0.2 M Solution :`M_1V_1 + M_2V_2 = M(V_1 + V_2)` ` 0.5 XX 800 + 1 xx 200 = M (800 + 200)` `M = (400+ 200)/(1000) = 0.6 M`

Discussion

28591.	The molarity of given orthophosphoric acid solution is 2M. Its normality is ………………….. .
Answer» 6N 4N 2N none of these Answer :A

Discussion

28592.	The major product obtained in the dehydrohalogenationof neo-pentyl bromide with alcoholic KOH is
Answer» 2-methylbut-1-ene 2,2-dimethylbut-1-ene 2-methylbut-2-ene but-2-ene. Solution :

Discussion

28593.	The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCI with 250 mL of 2(M) HCI will be
Answer» 0.975(M) 1.00(M) 0.875(M) 1.75(M) Solution :750mL 0.5(M) HCl contain `(0.5)/(1000)xx750=0.375mol` of HCl and 250mL of 2(M) HCl CONTAINS `2/(1000)xx250`=0.5 MOL of HCl. Volume of the resulting solution = 750 + 250 = 1000mL . Total NUMBER of moles of HCl in the solution = 0.375 + 0.5 = 0.875 mol `:.` Molarity of the resulting solution = 0.875(M)

Discussion

28594.	The major product obtained by the reaction of phenol and carbondioxide with NaOH and is
Answer» salicylaldehyde SALICYCLIC ACID phthalic acid BENZOIC acid Answer :B

Discussion

28595.	The major product is formed when toluene si reacted with chlorine in the presence of halogen carrier.
Answer» ANSWER :B

Discussion

28596.	The major product obtained by the following reactions is …..
Answer» SOLUTION :

Discussion

28597.	The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:
Answer» 1.75 M 0.975 M 0.875 M 1.00 M Solution :`M_(3)V_(3)=M_(1)V_(1)+M_(2)V_(2)` `M_(3)=((0.5M)xx(750mL)+(250 ML))/((750 mL+250 mL))` 0.875 M.

Discussion

28598.	The major product is
Answer» SOLUTION :

Discussion

28599.	The major product in the reaction of N-phenylbenzamide with Br_(2)//Fe is
Answer» ANSWER :B

Discussion

28600.	The molarity of a solution obtained by mixing 750 mL . Of 0.5 M Hcl with 250 mLof 2 M HCl will be
Answer» `0.875` M `1.00`M `1.75` M `0.0975` M ANSWER :a

Discussion

Explore topic-wise InterviewSolutions in .

The major product of the following reaction is-

The mole fraction of C_(2)H_(5)OH (Molar mass = 46 ) in 5 molal aqueous ethyl elchol solution is

The major product of the following reaction

The mole fraction of ethyl alcohol in its solution with methyl alcohol is 0.80. The vapour pressure of ethyl alcohol at the temperature of the solution is 40 mm of Hg. What is its vapour pressure in solution if the solution is ideal ?

The major product of the following reaction

The major product of the following reaction

The major product of the following addition reaction is :H_(3)C-CH=CH_(2) overset(Cl_(2)//H_(2)O) to

The mole fraction of benzene in a solution in toluene is 0.50. Calculate the weight percent of benzene in the solution.

The major product of the above reaction is

The mole fraction of a solute in a solutions is 0.1. At 298K molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm^(-3). The ratio of the molecular weights of the solute and solvent, (MW_("solute"))/(MW_("solvent")) is

The major product of the above reaction is

The mole fraction of a solute in a solution is 0.1 At 298K molarity of this solution is the same as its molality. The density of this solution at 298K is 2.0gcm^(-3). The ratio of the molecular weights of the solute and solvent, m_("solute")//m_("solvent") is ............

Themajor product of reaction is :

The mole fraction of a solute in its solution in acetic acid is 0.2. The mass of solute (molar mass = 40) in 120g. Of acetic acid would be

The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is same as its molality Density of the solution at 298 K is "2.0 g cm"^(-3). The ratio of the molecular weights of the solute and solvent ((MW_("solute"))/(MW_("solvent"))), is

The major product obtained when tert -butyl bromide is heated with sodium ethoxide is

The mole fraction of a solute in bezene solvent is 0.2. Themolality of the solution will be:

The major product obtained when tert-butyl bromide is heated with sodium ethoxide is

The mole fraction of a gas dissolved in a solvent is given by Henry's law. Constant for gas in water at 298 K is 5.55xx10^(7) Torr and the partial pressure of the gas is 200 Torr, then what is the amount of the gas dissolved in 1.0 kg of water?

The major product obtained when phenol is treated with sodium hydroxide and carbon di oxide is _______.

The mole fraction of a solute in a solution 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm^(-3). The ratio of the molecular weight of the solute and solvent, ((MW_("solute"))/(MW_("solvent"))), is

The major product obtained when, chlorobenzene is reacted with chlorine in the presence of FeCl_(3)

The major product obtained when aniline is treated with acetic anhydride in the the presence of pyridine is……….

The molarity of urea (molar mass 60 g mol^(-1)) solution by dissolving 15 g urea in 500 cm^(3) of water is

The major product obtained when 2-bromobutane is heated with alcoholic KOH is :

The major product obtained on treatment of CH_(3)CH_(2)CH(F)CH_(3) with CH_(3)O^(-)//CH_(3)OH is

The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is :

The major product obtained on treatement of CH_(3)CH_(2)CH(F)CH_(3)with CH_(3)O^(-)//CH_(3)OH is

The molarity of pure water (density of water=1g ml^-1)

the major product obtained on interaction of phenol with sodium hydroxide ad carbon dioxide is

What is the molality of pure water

The major product obtained in the photobromination of 2-methylbutane is

The major product obtained in the following reactions is

The molarity of orthophosphoric acid having purity of 70% by weight and specific gravity 1.54 would be

The molarity of HNO_(3) in a sample which has density 1.4 g/mL and mass percentage of 63% is …….

The majorproduct obtained in the photo catalysed bromination of 2-methylbutane is :

The molarity of a solution obtained by mixing 800 ml of 0.5 M HCl with 200 ml of 1 M HCl will be

The major product obtained in the following reaction is-

The major product obtained in the following reaction is :

The molarity of a solution obtained by mixing 800 mL of 0.5 M HCl with 200 mL of 1 M HCI will be

The molarity of given orthophosphoric acid solution is 2M. Its normality is ………………….. .

The major product obtained in the dehydrohalogenationof neo-pentyl bromide with alcoholic KOH is

The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCI with 250 mL of 2(M) HCI will be

The major product obtained by the reaction of phenol and carbondioxide with NaOH and is

The major product is formed when toluene si reacted with chlorine in the presence of halogen carrier.

The major product obtained by the following reactions is …..

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCl will be:

The major product is

The major product in the reaction of N-phenylbenzamide with Br_(2)//Fe is

The molarity of a solution obtained by mixing 750 mL . Of 0.5 M Hcl with 250 mLof 2 M HCl will be