QJEEM20Cha727: JEE Main 2020 Chemistry - Class 11 JEEM20Cha727 Solution: A 100 mL solution was made by adding 1.43 g of Na 2 CO 3 .xH 2…

Molar mass of Na₂CO₃.xH₂O

= 23 × 2 + 12 + 48 + 18x

= 46 + 12 + 48 + 18x

= (106 + 18x)

As n_factor in dissolution will be determined from net cationic or anionic charge; which is 2

Eq wt = ${M \over 2}$ = (53 + 9x)

volume = 100 ml = 0.1 Litre

Normality =

No. of equivalents of solute

Volume of solution (in L)

$\Rightarrow$ 0.1= ${{{{1.43} \over {53 + 9x}}} \over {0.1}}$

$\Rightarrow$ 53 + 9x = 143

$\Rightarrow$ 9x = 90

$\Rightarrow$ x = 10