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500 mL fo 2 M HCl, 100 mL of 2 M H_(2) SO_(4), and one gram equivalent of a monoacidic alkali are mixed together. 30mL of this solution requried 20 mL of 143 g Na_(2) CO_(3). xH_(2)O in one litre solution. Calculate the water of crystallisation of Na_(2) CO_(3). xH_(2) O

Answer» <html><body><p><br/></p>Solution :i. <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> mEq of acid `= 500 xx 2 + 100 xx 2 xx 2` (`n` factor) <br/> `= 1000 + 400 = 1400` <br/> Total mEq of monoacid alkali = 1000 <br/> Acid left after neutralisation with alkali <br/> `= 1400 - 1000 = 400 mEq` <br/> `N` fo acid left `= (400 mEq)/((500 + 100)mL) = (400)/(600) = 0.6 N` <br/> mEq of acid `-= mEq "of"Na_(2)CO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` <br/> `30 mL xx 0.66 N = 20 mL xx N Na_(2)CO_(3)` <br/> Strength of `Na_(2)CO_(3) x H_(2)O = 143 g L^(-1)` <br/> `M_(Na_(2)CO_(3).xH_(2)O) = ("Strenght")/(Mw) = ((143)/(106 + 18x))` <br/> `M_(Na_(2)CO_(3).xH_(2)O) = ((2 xx 143)/(106 + 18 x))` (`n` factor = 2) <br/> Substitution the value of `N` of `Na_(2)CO_(3).xH_(2)O` in <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (i), we get <br/> `30 xx 0.66 x = 20 xx ((2 xx 143)/(106 + 18 x))` <br/> solve for `x` <br/> `x = 10.16 ~~ 10` <br/> So <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> of crystallisation of `Na_(2)CO_(3).xH_(2) O = 10` <br/> formula : `Na_(2) CO_(3). 10 H_(2) O`</body></html>


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