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  • A 100mL solution was made by adding 1.43g of . The normality of the solution is 0.1N. The value of x is___________. (The atomic mass of

A 100mL solution was made by adding 1.43g of \mathrm{Na_{2}CO_{3}.xH_{2}O}. The normality of the solution is 0.1N. The value of x is___________. (The atomic mass of Na is 23 g/mol)
 

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Molar mass(M) of Na2CO3.xH2O
M =  23 × 2 + 12 + 48 + 18x
M =  46 + 12 + 48 + 18x
M  =  (106 + 18x)

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

Eq.wt. = \mathrm{\frac{M}{2}} = (53 + 9x)

\text { Gm}_\textup{eq }=\frac{\textup{wt}}{\text { Eq.wt. }}=\frac{1.43}{53+9 x}

\text { Normality }=\frac{\text { Gm}_\textup{eq} }{\mathrm{V}_{\text {litre }}}

As volume = 100 ml = 0.1 Litre

\text { Normality }=0.1=\frac{\frac{1.43}{53+9 \mathrm{x}}}{0.1}

10^{-2}=\frac{1.43}{53+9 x}

53 + 9x = 143
9x = 90
x = 10.00

Ans = 10

Posted by

Kuldeep Maurya

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