To a 25 ~mL H_2 O_2 solution, excess of acidified solution of KI was added. The iodine liberated required 20 ~mL of 0.3 ~N Na_2 ~S_2 O_3 solution. The volume strength of H_2 O_2 solution is

Correct answer: D. 4.08 ~g / L aligned & 2 KI+H_2 SO_4+H_2 O_2 K_2 SO_4+2 H_2 O+I_2 & 2 Na_2 ~S_2 O_3+I_2 Na_2 ~S_4 O_6+2 NaI aligned milli eq. of H_2 O_2 in 25 ~mL=20 × 0.3=6 milli eq. of H_2 O_2 in 1000 ~mL=(6)/(25) × 1000=240 Equivalent per litre = (240)/(1000) Gram per litre of H_2O_2=(240)/(1000) × 17=4.08 ~g / L (Equivalent weight of H_2 O_2=(34)/(2)=17 ).

AIIMS 2009

Exam: AIIMS
Year: 2009
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

AIIMS 2009 · Chemistry · Some Basic Concepts of Chemistry

To a 25 ~mL H_2 O_2 solution, excess of acidified solution of KI...

To a

25 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}_2

solution, excess of acidified solution of

\mathrm{KI}

was added. The iodine liberated required

20 \mathrm{~mL}

0.3 \mathrm{~N} \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3

solution. The volume strength of

\mathrm{H}_2 \mathrm{O}_2

solution is

Options

A $1.344 \mathrm{~g} / \mathrm{L}$
B $3.244 \mathrm{~g} / \mathrm{L}$
C $5.4 \mathrm{~g} / \mathrm{L}$
D $4.08 \mathrm{~g} / \mathrm{L}$

Correct answer

D. $4.08 \mathrm{~g} / \mathrm{L}$

Step-by-step solution

\begin{aligned} & 2 \mathrm{KI}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{I}_2 \\ & 2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3+\mathrm{I}_2 \longrightarrow \mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6+2 \mathrm{NaI}\end{aligned}

milli eq. of

\mathrm{H}_2 \mathrm{O}_2 in 25 \mathrm{~mL}=20 \times 0.3=6

milli eq. of

\mathrm{H}_2 \mathrm{O}_2 in 1000 \mathrm{~mL}=\frac{6}{25} \times 1000=240

Equivalent per litre =

\frac{240}{1000} Gram per litre

\mathrm{H}_2\mathrm{O}_2=\frac{240}{1000}

\times 17=4.08

\mathrm{~g}

\mathrm{L}

(Equivalent weight of

\mathrm{H}_2 \mathrm{O}_2=\frac{34}{2}=17

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