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Standard XIII

Chemistry

Balancing Redox Equations

2.48 g of Na...

Question

$2.48 g of N a_{2} S_{2} O_{3}$ . $x H_{2} O$ is dissolved per litre of solution. $20 mL$ of this solution required $10 mL$ of $0.01 M$ iodine solution for a complete reaction. What is the value of $x$ ?

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Solution

Complete reaction:
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
$(n - f a c t o r o f N a_{2} S_{2} O_{3}) = 1$
$(n - f a c t o r o f I_{2}) = 2$
$(n \times n_{f a c t o r})_{N a_{2} S_{2} O_{3}} = (n \times n_{f a c t o r})_{I_{2}}$
let M be molar mass of $N a_{2} S_{2} O_{3} . x H_{2} O$
$(\frac{2.48}{M}) \times 1 = 10 \times 2 \times 0.01$
$M = 248 g$
$2 (23) + 2 (32) + 3 (16) + x (18) = 248$
$x = 5$

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2.48 g of Na2S2O3. xH2O is dissolved per litre of solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution for a complete reaction. What is the value of x?

Complete reaction: 2Na2S2O3+I2→Na2S4O6+2NaI (n−factor of Na2S2O3)=1 (n−factor of I2)=2 (n×nfactor)Na2S2O3=(n×nfactor)I2 let M be molar mass of Na2S2O3.xH2O (2.48M)×1=10×2×0.01 M=248g 2(23)+2(32)+3(16)+x(18)=248 x=5

$2.48 g of N a_{2} S_{2} O_{3}$ . $x H_{2} O$ is dissolved per litre of solution. $20 mL$ of this solution required $10 mL$ of $0.01 M$ iodine solution for a complete reaction. What is the value of $x$ ?