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 %

EASY

JEE Main/Advance

IMPORTANT

Earn 100

$200 mL$ of a solution of mixture of $NaOH$ and ${Na}_{2} {CO}_{3}$ was first titrated with $Ph$ and $\frac{N}{10} HCl . 17.5 mL$ of $HCl$ was required for end point. After this $MeOH$ was added and $2.5 mL$ of same $HCl$ was again required for next end point. Find out amounts of $NaOH$ and ${Na}_{2} {CO}_{3}$ in the mix.

Important Questions on Redox Reactions

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 17

A solution contains

{Na}_{2} {CO}_{3}

and

{NaHCO}_{3} . 10 mL

of this requires

0.06 gm and 0.0265 gm

0.1 M H_{2} {SO}_{4}

for neutralisation using

Ph

indicator.

MeOH

is then added when a further

2.5 mL

0.2 M H_{2} {SO}_{4}

was needed. Calculate strength of

{Na}_{2} {CO}_{3}

and

{NaHCO}_{3}

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 6 - Redox Reactions>EXERCISE-4>Q 18

A sample containing

{Na}_{2} {CO}_{3} & NaOH

is dissolved in

100 mL

solution.

10 mL

of this solution requires

25 mL

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

%

{Na}_{2} {CO}_{3}

and

NaOH

in the sample.

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.

200 mL of a solution of mixture of NaOH and Na2CO3 was first titrated with Ph and N10HCl.17.5 mL of HCl was required for end point. After this MeOH was added and 2.5 mL of same HCl was again required for next end point. Find out amounts of NaOH and Na2CO3 in the mix.

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !