2.48 g of Na 2 S 2 O 3 .xH 2 O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.

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Chhavi Jain · Answer

Dear Student , We will use the formula : Molarity (M) = no. of moles of solute / volume of solution in Litres Na 2 S 2 O 3 .xH 2 O was dissolved in 1L of some solution. Now, 20 mL of this solution required 10mL of 0.01M iodine solution. no. of moles = M * V ( from the above formula ) n ( iodine solution)= 0.01 * 10*10 -3 = 10 -4 moles therefore , n ( solution) = 10 -4 = Molarity of solution * volume of solution 10 -4 = M sol * 20 * 10 -3 M sol = 0.005 Now, 2.48 g of Na 2 S 2 O 3 .xH 2 O is dissolved in 1L this solution . So, it is behaving as solute . Again using the same formula we will find the value of x : 0.005 = moles of solute / 1 L 0.005 = given mass / molar mass 0.005 = 2.48 / {46+64+48+x (18)} Now, solve this equation to get the value of x = ? Chemistry Consultant Askiitians

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2.48 g of Na 2 S 2 O 3 .xH 2 O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.

2.48 g of Na₂S₂O₃.xH₂O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.

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2.48 g of Na 2 S 2 O 3 .xH 2 O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.

2.48 g of Na2S2O3.xH2O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.

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2.48 g of Na₂S₂O₃.xH₂O was dissolved per litre of the solution. 20 mL of this solution required 10 mL of 0.01 M iodine solution. Find out the value of x.