Q: To form a complete monolayer of acetic acid on 1 ~g of charcoal, 100 ~mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of

aligned & Millimole of acid taken =100 0.5=50 \\ & Millimole of NaOH used =40 1=40 \\ & Millimole of acid adsorbed =50-40=10 \\ & Molecules of acid adsorbed =10 10^ -3 6 10^ 23 =6 10^ 21 \\ & Surface area occupied per molecule = 1.5 10^2 6 10^ 21 =0.25 10^ -19 =2500 10^ -23 aligned

Q: Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 ~g of an organic sal

Mass of salt produced = 1 . 29 g Moles of salt produced = 1 .29 129 = 0 .01 mole To produce, 0 . 01 mole organic salt 0 . 01 mole PhNO 2 and 0 . 03 mole SnCl 2 is required. 0 . 03 mole SnCl 2 is produced from 0 . 03 mole Sn . Mass of Sn = x gm = 0 .03 × 119 = 3 .57 g Mass of PhNO 2 = y gm = 0 .01 × 123 = 1 .23 g x = 3

Q: Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 ~g of an organic sal

Mass of salt produced = 1 . 29 g Moles of salt produced = 1 .29 129 = 0 .01 mole To produce, 0 . 01 mole organic salt 0 . 01 mole PhNO 2 and 0 . 03 mole SnCl 2 is required. 0 . 03 mole SnCl 2 is produced from 0 . 03 mole Sn . Mass of Sn = x gm = 0 .03 × 119 = 3 .57 g Mass of PhNO 2 = y gm = 0 .01 × 123 = 1 .23 g x = 3

Q: For the following reaction scheme, percentage yields are given along the arrow: x g and y g are mass of R and U , respectively. (Use: Molar mass (in g mol ^ -1 ) of H , C and O as 1,12 and 16 , respec

Mg 2 C 3 → H 2 O Mg ( OH ) 2 + C 3 H 4 ( P ) Propyne ( 4 g ) Now, one mole of propyne give, one mole of 2 -butyne. 3 moles of 2 -butyne give one mole of hexamethyl benzene. Hence, 40 g of propyne give 54 g to 2 -butyne 100 % but given that 75 % yeild and 4 g to propyne. The mass of 2 -butyne formed = 5 .4 × 75 100 = 4 .05 g 162 g t

Q: For the following reaction scheme, percentage yields are given along the arrow: x g and y g are mass of R and U , respectively. (Use: Molar mass (in g mol ^ -1 ) of H , C and O as 1,12 and 16 , respec

Mg 2 C 3 → H 2 O Mg ( OH ) 2 + C 3 H 4 ( P ) Propyne ( 4 g ) CH 3 − C ≡ CH → 33 K ( 100 % ) Hg 2 + CH 3 − C | | O − CH 3 ( S ) Now, one moe of propyne give one mole of acetone. 100 % . one mole of acetone give 0 . 5 moles of α , β -unsaturated compound 100 % yield . But given that 80 % yield. No. of moles of α , β -unsaturated compound = 0 .04 mol . = 0 .04

Q: Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters ( L ) produced at 300 K and 1 . 0 atm pressure, when 5

2 AI + 3 H 2 SO 4 ⟶ AI 2 SO 4 3 + 3 H 2 5 . 7 27 50 × 5 1000 = 0 . 2 mole = 0 .25 mole ( Limiting reagent ) ∴ 0 .25 mole H 2 is formed PV = nRT V = nRT P = 0 . 25 × 0 . 082 × 300 1 atm = 6 . 15 litre

Q: The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05 . If the density of the solution is 1.2 g c m - 3 , the molarity of urea solution is ______

Mole fraction ⇒ 0.05 Number of moles of water ⇒ 900 18 = 50 ∴ n u r e a n u r e a + 50 = 0.05 n u r e a = 2.63 Mass of urea = W u r e a = 2.63 × 60 ⇒ 157.8 g Total mass of solution ⇒ 157.8 + 900 ⇒ 1057.8 g Volume of solution ⇒ 1057.8 1.2 m l ⇒ 881.5 m l Molarity ⇒ n o . o f m o l e s o f s o l u t i o n s v o l u m e o f s o l u t i o n i n L ⇒

Question 1

To form a complete monolayer of acetic acid on 1 ~g of charcoal, 100 ~mL of 0.5 M acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 mL of

Accepted Answer

aligned & Millimole of acid taken =100 0.5=50 \ & Millimole of NaOH used =40 1=40 \ & Millimole of acid adsorbed =50-40=10 \ & Molecules of acid adsorbed =10 10^ -3 6 10^ 23 =6 10^ 21 \ & Surface area occupied per molecule = 1.5 10^2 6 10^ 21 =0.25 10^ -19 =2500 10^ -23 aligned

Question 2

To check the principle of multiple proportions, a series of pure binary compounds P m Q n  were analysed and their composition is tabulated below. The correct option(s) is(are) Compound Weight %

Accepted Answer

(A) If empirical formula of 3 is P 3 Q 4 , then molecular formula is P 3 Q 4 n &#8658; &#160;&#160; 3 M P 3 M P + 4 M Q = 40 100 = 2 5 15 M p = 6 M p + 8 M Q 9 M P = 8 M Q &#8756; For P 3 Q 5 , % of P = 3 M P 3 M P + 5 M Q &#215; 100 = 8 M Q 23 M Q &#215; 100 = 34 . 78 % (B) If empirical formula of compound 3 is P 3 Q 2 , we have 3 M P 3 M P + 2 M Q = 2 5 15 M P = 6 M P + 4 M Q 9 Mp = 4 M Q If M p = 20 M Q = 180 4 = 45 (C) If empirical formula of 2 is PQ &#8756; &#160;&#160; M P M P + M Q =

Question 3

The treatment of an aqueous solution of 3 . 74 &#160; g of Cu NO 3 2 with excess KI results in a brown solution along with the formation of a precipitate. Passing H 2 S through this brown solution giv

Accepted Answer

Number of moles of  Cu NO 3 2 = 3 . 74 187 = 0 . 02 2 Cu NO 3 2 + 4 KI &#8594; Cu 2 I 2 &#8595; + I 2 + 4 KNO 3 Number of moles of Cu 2 l 2 precipitated  = 0 . 01 I 2 brown &#160; solution + H 2 S &#8594; S &#8595; + 2 HI Number of moles of S precipitated  = 0 . 01 Mass of S precipitates  = 0 . 01 &#215; 32 g = 0 . 32 &#160; g

Question 4

Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 ~g of an organic sal

Accepted Answer

Mass of salt produced  = 1 . 29 &#160; g Moles of salt produced  = 1 .29 129 = 0 .01 &#160; mole To produce,  0 . 01 &#160; mole  organic salt  0 . 01 &#160; mole PhNO 2  and  0 . 03 &#160; mole &#160; SnCl 2  is required. 0 . 03 &#160; mole &#160; SnCl 2  is produced from  0 . 03 &#160; mole   Sn . Mass of  Sn = x &#160; gm = 0 .03 &#215; 119 = 3 .57 &#160; g Mass of  PhNO 2 = y &#160; gm = 0 .01 &#215; 123 = 1 .23 &#160; g x = 3

Question 5

Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 ~g of an organic sal

Accepted Answer

Mass of salt produced  = 1 . 29 &#160; g Moles of salt produced  = 1 .29 129 = 0 .01 &#160; mole To produce,  0 . 01 &#160; mole  organic salt  0 . 01 &#160; mole PhNO 2  and  0 . 03 &#160; mole &#160; SnCl 2  is required. 0 . 03 &#160; mole &#160; SnCl 2  is produced from  0 . 03 &#160; mole   Sn . Mass of  Sn = x &#160; gm = 0 .03 &#215; 119 = 3 .57 &#160; g Mass of  PhNO 2 = y &#160; gm = 0 .01 &#215; 123 = 1 .23 &#160; g x = 3

Question 6

For the following reaction scheme, percentage yields are given along the arrow: x g and y g are mass of R and U , respectively. (Use: Molar mass (in g mol ^ -1 ) of H , C and O as 1,12 and 16 , respec

Accepted Answer

Mg 2 C 3 &#8594; H 2 O Mg ( OH ) 2 + C 3 H 4 ( P ) Propyne &#160; ( 4 g ) Now, one mole of propyne give, one mole of  2 -butyne.  3  moles of  2 -butyne give one mole of hexamethyl benzene. Hence,  40 &#160; g  of propyne give  54 &#160; g  to  2 -butyne  100 % but given that  75 %  yeild and  4 &#160; g  to propyne. The mass of  2 -butyne formed  = 5 .4 &#215; 75 100 = 4 .05 &#160; g   162 &#160; g  t

Question 7

For the following reaction scheme, percentage yields are given along the arrow: x g and y g are mass of R and U , respectively. (Use: Molar mass (in g mol ^ -1 ) of H , C and O as 1,12 and 16 , respec

Accepted Answer

Mg 2 C 3 &#8594; H 2 O Mg ( OH ) 2 + C 3 H 4 ( P ) Propyne &#160; ( 4 g ) CH 3 &#8722; C &#8801; CH &#8594; 33 K ( 100 % ) Hg 2 + CH 3 &#8722; C | | O &#8722; CH 3 ( S ) Now, one moe of propyne give one mole of acetone.  100 % . one mole of acetone give  0 . 5  moles of  &#945; , &#946; -unsaturated compound  100 % &#160; yield . But given that  80 %  yield. No. of moles of  &#945; , &#946; -unsaturated compound  = 0 .04 &#160; mol .  = 0 .04

Question 8

Consider the reaction sequence from P  to Q shown below. The overall yield of the major product Q from P 75 % . What is the amount in grams of Q obtained from 9 . 3 &#160; mL P ? (Use densit

Accepted Answer

⇒ Molar mass of ' P ' C 6 H 7 N = 93 g ⇒ Mole of ' P ' = 9 .3 g ml × 1 ml × 1 93 g = 0 .1 mole ⇒ Molar mass of ' Q ' = 248 g ⇒ Mole of ' Q ' produced = 0 . 1 mole Mass of ' Q ' produced = ( 0 .1 × 248 × 0 .75 ) g = 18 .60 g

Question 9

Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters ( L )  produced at  300 &#160; K and 1 . 0 atm pressure, when  5

Accepted Answer

2 AI + &#160; 3 H 2 SO 4 &#10230; AI 2 SO 4 3 + 3 H 2 5 . 7 27   50 &#215; 5 1000 = 0 . 2 &#160; mole   = 0 .25 &#8201; mole ( Limiting &#8201; reagent ) &#8756;   0 .25 &#8201; mole &#8201;H 2 is formed PV = nRT V = nRT P = 0 . 25 &#215; 0 . 082 &#215; 300 1 &#160; atm = 6 . 15  litre

Question 10

The mole fraction of urea in an aqueous urea solution containing 900 &#160; g of water is 0.05 . If the density of the solution is 1.2 &#160; g &#160; c m - 3 , the molarity of urea solution is ______

Accepted Answer

Mole fraction &#8658; 0.05 Number of moles of water &#8658; 900 18 = 50 &#8756; n u r e a n u r e a + 50 = 0.05 n u r e a = 2.63 Mass of urea = W u r e a = 2.63 &#215; 60 &#8658; 157.8 &#8201; g Total mass of solution &#8658; 157.8 + 900 &#8658; 1057.8 &#8201; g Volume of solution &#8658; 1057.8 1.2 m l &#8658; 881.5 &#8201; m l Molarity &#8658; n o . &#8201; o f &#8201; m o l e s &#8201; o f &#8201; s o l u t i o n s v o l u m e &#8201; o f &#8201; s o l u t i o n &#8201; i n &#8201; L &#8658;

Some Basic Concepts of Chemistry — Previous Year Questions & Solutions