0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the platinum residue on ignition. Calaculate the molecular mass of the base.

0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the pl

1.	0.532 g of the chloroplatinate of a diacid base gave 0.195 g of the platinum residue on ignition. Calaculate the molecular mass of the base.
Answer» <html><body><p></p>Solution :Mass of the base chloroplatinate taken `(W) = 0.532 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>` <br/> Mass of the platinum residue `(x) = 0.<a href="https://interviewquestions.tuteehub.com/tag/195-281683" style="font-weight:bold;" target="_blank" title="Click to know more about 195">195</a> g` <br/> step I. Equivalent mass of the base <br/> Equivalent mass of the base `= (1)/(2) [(W)/(x) xx 195 - 410] = (1)/(2) [(0.532)/(0.195) xx 195 - 410]` <br/> `= (1)/(2) [532 - 410] = <a href="https://interviewquestions.tuteehub.com/tag/61-330126" style="font-weight:bold;" target="_blank" title="Click to know more about 61">61</a>` <br/> Step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Molecular mass of the base <br/> Molecular mass of the base = Equivalent mass `xx` <a href="https://interviewquestions.tuteehub.com/tag/acidity-847681" style="font-weight:bold;" target="_blank" title="Click to know more about ACIDITY">ACIDITY</a> `= 61 xx 2 = 122`</body></html>