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Standard XII

Chemistry

Types of Redox Reactions

When 25 ml of...

Question

When 25 ml of $H_{2} O_{2}$ solution having volume strength of 4.48, is treated with borax in presence of NaOH solution, the no. of millimoles of sodium perborate formed is:

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Solution

Molar concentration of $H_{2} O_{2}$ solution of volume strength $4.48 = \frac{4.48}{11.2} = 0.4 M$
So, mmols of $H_{2} O_{2}$ in 25 ml solution $= 0.4 \times 25$ mmols
$N a_{2} B_{4} O_{7} + 4 H_{2} O_{2} + 2 N a O H \to 2 N a_{2} B_{2} O_{4} (O H)_{4} + H_{2} O$
So, mmols of sodium perborate formed = 5 mmols

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When 25 ml of H2O2 solution having volume strength of 4.48, is treated with borax in presence of NaOH solution, the no. of millimoles of sodium perborate formed is:

Molar concentration of H2O2 solution of volume strength 4.48=4.4811.2=0.4M So, mmols of H2O2 in 25 ml solution =0.4×25 mmols Na2B4O7+4H2O2+2NaOH→2Na2B2O4(OH)4+H2O So, mmols of sodium perborate formed = 5 mmols

When 25 ml of $H_{2} O_{2}$ solution having volume strength of 4.48, is treated with borax in presence of NaOH solution, the no. of millimoles of sodium perborate formed is: