2.86 g of Na2CO3.xH2O in 100 ml solution is 0.2 N. Then the value of x is,

molar mass of Na2co3.xH2o=(106+18x)2.86 g salt = (2.86106+18x) mole in 100mL Molarity = (28.6106+18x) mole in 1000mLFor Na2Co3, normality=2Xmolarity=0.2N (2X28.6106+18x) =0.2 x=10

2.86 g of Na2CO3.xH2O in 100 ml solution is 0.2 N. Then the value of x is,

molar mass of Na2co3.xH2o=(106+18x)2.86 g salt = (2.86106+18x) mole in 100mL Molarity = (28.6106+18x) mole in 1000mLFor Na2Co3, normality=2Xmolarity=0.2N (2X28.6106+18x) =0.2 x=10

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2.86 g of Na₂CO₃.xH₂O in 100 ml solution is 0.2 N. Then the value of x is,

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answer is B.

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Detailed Solution

$\begin{array}{l} {molar mass of Na}_{2} c o_{3} . x H_{2} o = (106 + 18 x) \\ 2.86 g salt = (\frac{2.86}{106 + 18 x}) mole in 100mL \\ Molarity = (\frac{28.6}{106 + 18 x}) mole in 1000mL \\ F o r {Na}_{2} C o_{3}, normality=2Xmolarity=0.2N \\ (\frac{2 X 28.6}{106 + 18 x}) = 0.2 \\ x = 10 \end{array}$