What amount of substance containing 60 % NaCl , 37 % KCl should be weighed out for analysis so that after the action of 25 mL of 0 . 1 NAgNO 3 solution, excess of Ag + is back titrates with 5 mL of NH 4 SCN solution. Given that 1 mL of NH 4 SCN = 1 . 1 . mL of AgNO 3 .

Let the mass of sample = x g 60 % NaCl : - w NaCl = 60 × x 100 = 0 . 6 x g 37 % KCl : - W KCl = 37 × x 100 = 0 . 37 x g So, m eq of silver nitrate = 25 × 0 . 1 = 2 . 5 m eq of NH 4 SCN = m eq of AgNO 3 N × 1 = 0 . 1 × 1 . 1 N = 0 . 11 m eq of AgNO 3 left = m eq of NH 4 SCN m eq of AgNO 3 left = 5 × 0 . 11 = 0 . 55 So, m eq of AgNO 3 = 2 . 5 - 0 . 55 = 1 . 95 m eq of KCl + m eq of NaCl = 1 . 95 0 . 6 x 58 . 5 + 0 . 37 x 74 . 5 × 1000 = 1 . 95 So, x = 0 . 1281 gm .

1 . 16 gCH 3 CH 2 COOH was bumt in excess air and the resultant gases CO 2 and H 2 O were passed through excess NaOH solution. The resulting solution was divided in two equal parts. One part required 50 mL of 1 NHCl for neutralization using phenolphthalein as indicator. Another part required 80 mL of 1 N HCl for neutralization using methyl orange as indicator. Find the value of n and the amount of excess NaOH solution taken initially.

Given: Weight of CH 3 CH 2 COOH = 1 . 16 gm Two parts : 50 mL of 1 N HCl and 80 mL of 1 N HCl CH 3 CH 2 n COOH + O 2 ⟶ n + 2 CO 2 + H 2 O 1 mole of acid = n + 2 moles of CO 2 Let moles of acid be a , So moles of CO 2 = a n + 2 CO 2 ( n + 2 ) a + 2 NaOH b ⟶ Na 2 CO 3 (Let the moles of NaOH be b) Moles of NaOH left = b - ( n + 2 ) a Moles of Na 2 CO 3 = ( n + 2 ) a moles = mass atomic mass m . eq = moles × n - factor On dividing into equal parts moles get halved. Part-I (With phenolphthalein)undefined m . eq of Na 2 CO 3 + m . eq of NaOH left = m . eq of HCl b - 2 ( n + 2 ) a 2 + ( n + 2 ) a 2 = 0 . 05 . . . i Part-II (With methyl orange) m . eq of Na 2 CO 3 + m . eq of NaOH left = m . eq of HCl b - 2 ( n + 2 ) a 2 + 2 ( n + 2 ) a 2 = 0 . 08 . . . ii By i and ii ( n + 2 ) a 2 = 0 . 03 ( n + 2 ) a = 0 . 06 . . . iii and 1 . 16 60 + 14 n = a . . . iv from equation ( iii ) & ( iv ) 1 . 16 60 + 14 n = 0 . 06 ( n + 2 ) 19 . 33 n + 38 . 66 = 60 + 14 n 5 . 33 n = 21 . 33 ⇒ n = 4 from equation iii ⇒ 6 a = 0 . 06 ⇒ a = 0 . 01 from equation i ⇒ b 2 - 0 . 06 + 0 . 03 = 0 . 05 ⇒ b 2 = 0 . 08 moles of NaOH ⇒ b = 0 . 16 Mass of NaOH = 0 . 16 × 40 = 6 . 4 g Hence, the value of n is 4 and amount of NaOH solution taken is 6 . 4 g

What volume at NTP of gaseous ammonia will be required to be passed into 30 cm 3 of 1 N H 2 SO 4 solution to bring down the acid strength of the latter to 0 . 2 N ?

Milli-equivalent of H 2 SO 4 reacted = 30 × 1 - 30 × 0 . 2 = 24 Milli-equivalent of NH 3 reacted = milli equivalent of H 2 SO 4 reacted = 24 'm' moles of NH 3 = milli equivalent of NH 3 × n -factor = 24 × 1 = 24 Volume of NH 3 at STP = 24 × 22 . 4 = 537 . 6 mL

0 . 5 g of fuming H 2 SO 4 (oleum) is diluted with water. The solution requires 26 . 7 mL of 0 . 4 N NaOH for complete neutralization. Find the % of free SO 3 in the sample of oleum.

Given: Weight of H 2 SO 4 = 0 . 5 g , 26 . 7 mL of 0 . 4 N NaOH Oleum consists of SO 3 and H 2 SO 4 Let the mass of SO 3 in the given sample of oleum be = x g Mass of H 2 SO 4 in the given sample of oleum = ( 0 . 5 − x ) g 2 NaOH + SO 3 → Na 2 SO 4 + H 2 O 2 NaOH + H 2 SO 4 → Na 2 SO 4 + 2 H 2 O gm eq . = mass equivalent mass gm eq . = Normality × volume ( L ) No. of gm eq. of SO 3 = x 40 No. of gm eq. of H 2 SO 4 = 0 . 5 - x 49 Total gm eq. of oleum = x 40 + 0 . 5 - x 49 By the law of equivalence: gm eq. of oleum =gm eq. of NaOH x 40 + 0 . 5 - x 49 = 0 . 4 × 26 . 7 1000 X = 0 . 9328 9 = 0 . 1036 % of SO 3 = 0 . 1036 0 . 5 × 100 = 20 . 72 % Hence, % of SO 3 is 20 . 72 %

A volume of 12 . 53 mL of 0 . 05093 M SeO 2 reacted with exactly 25 . 52 mL of 0 . 1 M CrSO 4 . In the reaction, Cr 2 + was oxidized to Cr 3 + . To what oxidation state was selenium converted by the reaction.

Given: Volume of 0 . 05093 M SeO 2 = 12 . 53 mL , Volume of 0 . 1 M CrSO 4 = 25 . 52 mL Se + 4 O 2 + Cr + 2 SO 4 ⟶ Cr + 3 + Se + n By the law of equivalence: Meq . of SeO 2 = Meq . of CrSO 4 n --factor of Se = n - 4 n -factor of Cr = 3 - 2 = 1 Meq . = molality × n - factor × volume ⇒ 12 . 53 × 0 . 05093 ( n - 4 ) = 25 . 52 × 0 . 1 × 1 ⇒ n - 4 = 3 . 999041 ⇒ n ≃ 0 Hence, selenium is converted from Se 4 + to Se

EASY

JEE Main/Advance

IMPORTANT

Earn 100

A sample containing ${Na}_{2} {CO}_{3} & NaOH$ is dissolved in $100 mL$ solution. $10 mL$ of this solution requires $25 mL$ of $0.1 N HCl$ when $Ph$ is used as indicator. If $MeOH$ is used as indicator $10 mL$ of same solution requires $30 mL$ of same $HCl$ . Calculate $%$ of ${Na}_{2} {CO}_{3}$ and $NaOH$ in the sample.

Important Questions on Redox Reactions

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 19

Potassium acid oxalate

K_{2} C_{2} O_{4} \cdot 3 H_{2} C_{2} O_{4} \cdot 4 H_{2} O

can be oxidized by

{MnO}_{4}^{-}

in acid medium. Calculate the volume of

0.1 M {KMnO}_{4}

reacting in acid solution with one gram of the acid oxalate.

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 20

What amount of substance containing

60 % NaCl, 37 % KCl

should be weighed out for analysis so that after the action of

25 mL

0.1 {NAgNO}_{3}

solution, excess of

{Ag}^{+}

is back titrates with

5 mL

{NH}_{4} SCN

solution. Given that

1 mL

{NH}_{4} SCN = 1.1 . mL

{AgNO}_{3}

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 21

25 mL

of a solution containing

HCl

was treated with excess of

\frac{M}{5 {KIO}_{3}}

and

KI

solution of unknown concentration where

I_{2}

liberated is titrated against a standard solution of

0.021 M {Na}_{2} S_{2} O_{3}

solution whose

24 mL

were used up. Find the strength of

HCl

and volume of

{KIO}_{3}

solution consumed :

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 22

1.16 {gCH}_{3} ({CH}_{2}) COOH

was bumt in excess air and the resultant gases

({CO}_{2}

and

H_{2} O)

were passed through excess

NaOH

solution. The resulting solution was divided in two equal parts. One part required

50 mL

1 NHCl

for neutralization using phenolphthalein as indicator. Another part required

80 mL

1 N

HCl

for neutralization using methyl orange as indicator. Find the value of

n

and the amount of excess

NaOH

solution taken initially.

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 24

What volume at

NTP

of gaseous ammonia will be required to be passed into

30 {cm}^{3}

1 N H_{2} {SO}_{4}

solution to bring down the acid strength of the latter to

0.2 N

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 25

0.5 g

of fuming

H_{2} {SO}_{4}

(oleum) is diluted with water. The solution requires

26.7 mL

0.4 N NaOH

for complete neutralization. Find the

%

of free

{SO}_{3}

in the sample of oleum.

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 26

A volume of

12.53 mL

0.05093 M {SeO}_{2}

reacted with exactly

25.52 mL

0.1 M {CrSO}_{4}

. In the reaction,

{Cr}^{2 +}

was oxidized to

{Cr}^{3 +}

. To what oxidation state was selenium converted by the reaction.

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 27

Metallic tin in presence of

HCl

is oxidised by

K_{2} {Cr}_{2} O_{7}

to stannic chloride. What volume of decinormal dichromate solution will be reduced by

1 g

of tin ? At. wt. of

Sn =

118.7

A sample containing Na2CO3 & NaOH is dissolved in 100 mL solution. 10 mL of this solution requires 25 mL of 0.1 N HCl when Ph is used as indicator. If MeOH is used as indicator 10 mL of same solution requires 30 mL of same HCl. Calculate % of Na2CO3 and NaOH in the sample.

Important Questions on Redox Reactions

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