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1 grams of a carbonate (M_(2)CO_(3))on tratment with excess HCl produces 0.01186 mole of CO_(2). The molar mass of M_(2)CO_(3)in gmol^(-1) is .....

Answer» <html><body><p>1186<br/>84.3<br/>118.6<br/>11.86</p>Solution :`M_(2)CO_(3) + 2HCl rarr MCl_(2) + H_(2)O + CO_(2)` <br/> `(1)/(M_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>))` <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> `rarr ""0.01186 ` mole <br/> where , `M_(0) = ` <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> mass of `M_(2)CO_(3)` <br/> `(1)/(M_(0))= 0.01186` <br/> `:.M_(0)=84.3` gram/mole</body></html>


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