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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 28601. |
The major product in the reaction of 2-butyne will Li/Liq.NH_3 is |
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| 28602. |
The major product in the reaction is ……………….. {:(""CHO),("|"),(" "(CHOH)_(4)),("|"),(""CH_(2)OH):}overset((i)HCN)underset((ii)H_(3)O^(+))rarroverset(HI)rarr |
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Answer» Heptanoic ACID |
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| 28603. |
The molarity of a solution obtained by mixing750 mL of 0.5 (M) HCl with 250 mL of 2(M) HCI will be: |
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Answer» 0.875 M `M_("final") = (0.5 xx 750 + 2xx 250)/(1000)` ` = (375 + 500)/(1000) = 0.875 M` |
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| 28604. |
The major product in the reaction is ……. |
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| 28605. |
The major product in the reaction is …… |
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| 28606. |
The major product in the reaction is ……. |
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| 28607. |
The major product in the reaction, CH_(3)CH_(2)NH_(2)+Cl-overset(O)overset(||)C-CH_(3) is |
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Answer» N - Methylaminoethane |
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| 28608. |
The major product in the given reaction is ….. CH_(3)-CH_(2)-underset("I")underset("|")("C")H-CH_(2)-CH_(3)overset(NaOH)underset(H_(2)O)rarr |
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Answer» `CH_(3)-underset("OH ")underset("| ")("C ")H-CH_(2)-CH_(2)CH_(3)` 
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| 28609. |
The molarity of a NaOH solution by dissolving 4 g of it in 250 ml water is |
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Answer» 0.4 M |
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| 28610. |
The major product in the reaction : |
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Answer» a HEMIACETAL |
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| 28611. |
The molarity of a glucose solution containing 36 g of glucose per 400 ml of the solution is : |
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Answer» 1 |
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| 28612. |
The major product in the given reaction will be ….. CH_(3)-overset("Br ")overset("| ")("C ")H-CH_(2)CH_(3)+CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C ")-O^(-)K^(+)rarr |
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Answer» `CH_(3)-CH=CH-CH_(3)` |
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| 28613. |
Calculate the molality of a 9.8% (w/w) solution of H_2SO_4. |
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Answer» 1M |
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| 28614. |
The major product in the following reaction is having how many pi electrons here ? |
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| 28615. |
The molarity of a 100 mL solution containing 5.1 g of hydrogen peroxide is: |
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Answer» 0.15 M |
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| 28616. |
The major product in the following reaction is CH_(3)-underset(H)underset(|)overset(CH_(3))overset(|)(C)-CH_(2)Br underset(CH_(3)OH)overset(CH_(3)O^(-))to |
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Answer» `CH_(3)-UNDERSET(H)underset(|)overset(CH_(3))overset(|)(C)-CH_(2)OCH_(3)` 
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| 28618. |
The majorproduct in the following reaction is : |
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| 28619. |
The molarity of a 10% NaOH solution is |
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Answer» 2.5 `"Molarity"= ("Mass"//dm^(3))/"MOLAR mass"=100/40=2.5 M` |
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| 28620. |
The major product in the following reaction is : |
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| 28621. |
The major product from the reaction of Br_(2) with Z-3-hexene is: |
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Answer» optically ACTIVE racemic MIXTURE |
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| 28622. |
The molarity of 900 g of water is |
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Answer» 50 M `"900 g "H_(2)O=(900)/(18)"moles = 50 moles"` `"900 g"H_(2)O="900 ML of "H_(2)O` `THEREFORE"Molarity "=(50)/(900)xx100x=55.5M` |
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| 28623. |
The major product 'H' of the given reaction sequence is CH_(3)-CH_(2)-CO-CH_(3) overset(.^(Theta)CN)to G underset("Heat")overset(95%H_(2)SO_(4))toH |
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Answer» `CH_(3)-CH=UNDERSET(CH_(3))underset(|)(C)-COOH` Please note that hydrolysis of cyanides to carboxylic acids requries addition of amolecule of `H_(2)O`.95% `H_(2)SO_(4)` cannot SUPPLY `H_(2)O`, therefore, dehydration of (G) occurs to give (H). |
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| 28624. |
The molarity of 98% H_2SO_4(d = 1.8 g/mL) by weight is |
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Answer» 6 M volume of 100 g solution ` = (100)/(1.8) = 55.55 ml ` Molarity`= ("number of MOLES of " H_2SO_4)/("volume of solution ") xx 1000` ` = (98)/(98 xx 55.55) xx 1000 = 18 M ` |
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| 28625. |
The major product H in the given reaction sequence is CH_(3)-CH_(2)-CO-CH_(3)overset(.^(Theta)CN)rarr G overset(95% H_(2)SO_(4))underset("Heat")rarr H |
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Answer» `CH_(3)-CH=underset(" "CH_(3))underset(|)(C )-COOH` 
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| 28626. |
The molarity of 90% H_(2)SO_(4) solution is [density = 1.8 gm/ml] |
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Answer» `1.8` Weight of one litre of solution = 1800 gm `therefore` Weight of `H_(2)SO_(4)` in the solution `=(1800xx90)/(100)=1620 gm` `therefore` Weight of solvent = 1800 - 1620 = 180 gm `therefore` MOLALITY `=(1620)/(98)XX(1000)/(180)=91.83` |
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| 28627. |
The major product in the acid catalysed dehydration of 2-pentanol is : |
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Answer» 4-pentene |
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| 28628. |
The major product formed when aniline reacts with concentrated H_2SO_4 followed by heating with H_2SO_4 at 453-473k is |
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| 28629. |
The molarity and molality of a solution are M and m respectively. If the molecular weight of the solute is M', calculate the density of the solution in terms of M, m and M' |
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| 28630. |
The major product formed when alcoholic AgNO_2 reacts with ethyl chloride is |
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Answer» ETHYL NITRITE |
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| 28631. |
The molarity of 0.006 mole of NaCl in 100ml solution is |
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Answer» `0.6` |
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| 28632. |
The major product formed when 1, 1, 1-trichloro-propane is treated with aqueous potassium hydroxide is: |
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Answer» Propyne |
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| 28633. |
The molar volumes of ice and water are respectively 0.0196 and 0.0180 litres per mole at 273 K. If DeltaH for the transition of ice to water is 1440 calories per mole at 1 atm pressure, find Delta u. |
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| 28634. |
The major product formed in the reaction is (AAK_MCP_35_NEET_CHE_E35_014_Q01) |
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| 28635. |
The molar volume of liquid benzene (density =0.877g//mL) increases by a factor of 2750 as it vaporises at 20^(@)C and that of liquid toluene (density =0.867g//mL) increases by a factor of 7720 at 20^(@)C. A solution of benzene and toluene at 20^(@)C has a vapour pressure of 46 torr. Find the mole fraction of benzene in the vapour above the solution |
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Answer» <P> Solution :The volume of 1 mole of `C_(6)H_(6)` (liquid) `=(78)/(0.877)mL`The volume of 1 mole of `C_(6)H_(6)` (VAPOUR) `=(78)/(0.877)xx2750mL` `=244.58` LITRES for 1 mole of `C_(6)H_(6)` vapour at `20^(@)C` `p_(b)^(0)V=RT` `p_(b)^(0)=(0.0821xx293)/(244.58)`atm`=74.75mm` Similarly for 1 mole of `C_(6)H_(5)CH_(3)` vapour at `20^(@)C` `p_(t)^(0)=(0.0821xx293)/(((92)/(0.867)xx(7720)/(1000)))"atm"=22.32mm` Now, `(1)/(p)=(x_(b).)/(p_(b)^(0))+((1-x_(b).)/(p_(t)^(0)))` `(1)/(46)=(x_(b).)/(74.75)+(1-x_(b).)/(22.32)` `x_(b)^(.)=0.7336` |
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| 28636. |
The major product formed in the reaction is, Et – S-CH_2 -CH(CI)CH_3 overset( aq.KOH )(to) |
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| 28637. |
The molar volume of methane , CH_(4). At 819^(@)C and 760mm pressure is : |
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Answer» <P>22.4L `p_(1) = 760 mm HG ` `T_(1) = 273K` `V_(2) = ?` `p _(2) `= 760mm Hg `T_(2) = ` 1092 K `( V_(2))/( T_(2)) = ( V_(1))/( T_(1))` or `V_(2) = ( V_(1) T_(2))/( T_(1)) = ( 22.4 xx 1092)/( 273) = 89.6L` |
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| 28638. |
The major product formed on nitration of N, N - dimethylaniline with conc. H_2SO_4,NHO_3 mixture is |
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| 28640. |
The molar volume of liquid benzene (density =0.877 g ml^(-1)) increases by a factor of 2750 of it vaporises at 20^(@)C At 27^(@)C when a non-votalite solute (that does not to be 98.88 mm Hg. Calculate the freezing point of the solution. Given Enthalphy f vaporization of benzene (f) 394.57 J g^(-1) Enthalpy of fusion constant for benzne =5.0 kg "mol"^(-1) |
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Answer» SOLUTION :LET moles of benzene vaproises, at `20^(@)C=n_(1)` volume of `n_(1)` mole of benzene `(f)=(78n_(1))/(0.877)` Volume of `n_(1)` mole of benzene `(g)=2750 ((78n_(1))/(0.877))` `PV=nRT` `P^(@)=((78n_(1))/(0.877))((2750)/(1000))=n_(1)xx0.082xx293` `P_("Benzene")=0.0982` atm. `=74.63 mm Hg` `P_("Benzene")^(@)` at `27^(@)C` can be CALCULATED as `log((P_(2))/(P_(1)))=(DeltaH_("vap"))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `log ((P_(2))/(74.63))=(394.57xx78)/(2.303xx8.314)[(7)/(300xx293)]` `P_("Benzene")^(@) ("at"27^(@)C)=100.2` mm Hg Molality of the solution `=(X_("solvent")xx1000)/(X_("solvent")xx78)=(((100.2-98.88)/(98.88))xx10000)/(0.98xx78)=0.17` `Deltat_(f)=K_(f)m=5xx0.17=0.85` We known that `K_(f), =[(RT_(f)^(2))/(1000DeltaH_(f))]M` `5=((8.314xxT_(1)^(2))/(1000xx10060))78, T_(f)-278.5K` |
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| 28641. |
The major product formed in the reaction: |
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| 28642. |
The major product formed in the reaction is : |
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| 28643. |
The molar volume of X(l)(d=0.9g/mL) increases by a factor of 3000 as it vaporises at 27^(@)C and that of Y(l)(d=0.88 g/mL) increases by a factor of 8000 at 27^(@). A miscible liquid solution of X and Y at 27^(@)C has a vapour pressure of 50 torr. The mole fraction of Y in solution is : ( Given : 0.082 atm/L/mol/K, Molar mass of X=75, Molar mass of Y=88) |
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Answer» 0.48 |
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| 28644. |
The major product formed in the folowing reaction is |
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| 28645. |
The major product formed in the following reaction is …… CH_(3)CH(Cl)CH_(2)-CH_(2)OH overset(" aq. KOH ")rarr |
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Answer» `CH_(3)CH=CHCH_(2)OH` |
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| 28646. |
The major product formed in the following reaction is : CH_3-overset(CH_(3))overset(|)underset(H)underset(|)C-CH_2Broverset(CH_3O^(-))underset(CH_3OH)to |
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Answer» `CH_3-overset(CH_(3))overset(|)underset(H)underset(|)C-CH_2OCH_3` |
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| 28647. |
The molar volume of liquid benzene (density = "0.877 g mL"^(-1)) increases by a factor of 2750 as it vaporises at 20^(@)C and that of liquid toluene (density "0.867 g mL"^(-1)) increases by a factor of 7720 at 20^(@)C. A solution of benzene and toluene at 20^(@)C has a vapour pressure of 46.0 torr. Find the the mole fraction of benzene in the vapour above the solution. |
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Answer» Solution :In vapour phase, `"1 MOLE of benzene, i.e., 78 g has volume at "20^(@)C=(78)/(0.877)xx"2750 mL"` `"1 mole of toluene, i.e., 92 g has volume at "20^(@)C=(92)/(0.867)xx"7720 mL"` `"Applying , PV = N RT"` `"For benzene vapour, "(P^(@))/(760)"(atm)"xx(78xx2750)/(0.877xx1000)(L)=1xx"0.0821 L atm K"^(-1)"mol"^(-1)xx293 K` `"or"P_(B)^(@)=74.74mm` `"For toluene vapour, "(P^(@))/(760)xx(92xx7720)/(0.867xx1000)=1xx0.0821xx293` `"or"P_(T)^(@)=22.37mm` `""P_("total")=x_(B)xxP_(B)^(@)+x_(T)xxP_(T)^(@)=x_(B)P_(B)^(@)+(1-x_(B))xxP_(T)^(@)` `THEREFORE""46=x_(B)=(74.74)+(1-x_(B))(22.37)` This on solving gives `x_(B)=0.45` (in the liquid phase) `therefore""x_(T)=1-0.45=0.55` (in the liquid phase) `"In vapour phase,"P_(B)=x_(B)P_(B)^(@)=0.45xx74.74=36.63mm` `""P_(T)=x_(T)P_(T)^(@)=0.55xx22.37=12.30 mm` `"Mole fraction of benzene in vapour phase "(y_(B))=(P_(B))/(P_(B)+P_(T))(36.63)/(36.63+12.30)=0.75` |
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| 28648. |
The major product formed in the following reaction is (AAK_MCP_37_NEET_CHE_E37_002_Q01) |
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Answer» 4-methoxy acetophenone |
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| 28649. |
The major product formed in the following reaction CH_(3)CH(Cl)CH_(2)-CH_(2)OH overset(Aq. KOH)tois |
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Answer» `CH_(3)CH=CH-CH_(2)OH` |
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| 28650. |
The major product (ester) of the following reaction is |
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Answer» A single STEREOISOMER (optically ACTIVE) |
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