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Byju's Answer

Standard VIII

Chemistry

Types of Solutions Based on the Concentration of the Solution

50 mL of an a...

Question

$50 m L$ of an aqueous solution of $H_{2} O_{2}$ was treated with an excess of $K I$ solution and dilute $H_{2} S O_{4}$ . The liberated iodine required $20 m L 0.1 N N a_{2} S_{2} O_{3}$ solution for complete interaction. Calculate the concentration of $H_{2} S O_{2}$ in $g / L$ .

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Solution

$H_{2} O_{2} + 2 K I + H_{2} S O_{4} \to K_{2} S O_{4} + 2 H_{2} O + I_{2}$
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
Eq. mass $H_{2} O_{2} = \frac{34}{2} = 17$
$20 m L 0.1 N N a_{2} S_{2} O_{3} = 20 m L 0.1 N I_{2}$ solution
$\equiv 20 m L 0.1 N H_{2} O_{2}$ solution
Amount of $H_{2} O_{2}$ in $50 m L$ aq. solution
$= \frac{0.1 \times 17}{1000} \times 20 = 0.034 g$
Concentration in $g / L = \frac{0.034}{50} \times 1000 = 0.68$ .

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were treated with an excess of

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H_{2} S O_{4}

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Q. A

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of the diluted solution after complete extraction of iodine requires

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Q. An aqueous solution containing

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50 mL of an aqueous solution of H2O2 was treated with an excess of KI solution and dilute H2SO4. The liberated iodine required 20 mL 0.1 N Na2S2O3 solution for complete interaction. Calculate the concentration of H2SO2 in g/L.

H2O2+2KI+H2SO4→K2SO4+2H2O+I22Na2S2O3+I2→Na2S4O6+2NaIEq. mass H2O2=342=1720 mL 0.1 N Na2S2O3=20 mL 0.1 N I2 solution≡20 mL 0.1 N H2O2 solutionAmount of H2O2 in 50 mL aq. solution=0.1×171000×20=0.034 gConcentration in g/L=0.03450×1000=0.68.

$50 m L$ of an aqueous solution of $H_{2} O_{2}$ was treated with an excess of $K I$ solution and dilute $H_{2} S O_{4}$ . The liberated iodine required $20 m L 0.1 N N a_{2} S_{2} O_{3}$ solution for complete interaction. Calculate the concentration of $H_{2} S O_{2}$ in $g / L$ .