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60 gr of oleum (labelled as 118%) is mixed with 11.8 gm of water. What will be that composition of final mixture?A. Only `H_(2)SO_(4)`, having mass 71.8 gmB. 118 gm of `H_(2)SO_(4)`C. 70.8 gm `H_(2)SO_(4)` & 1 gm waterD. 32 gm `SO_(3)` & 39.8 gm `H_(2)SO_(4)` |
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Answer» Correct Answer - 3 100 gm will contain 80 gm `SO_(3)` 60 gm will contain `80/100xx60=48 gm ` `{:(SO_(3),+,H_(2)O,to,H_(2)SO_(4)),(48 gm,,10.8 gm ,,58.8 gm),(0.6"mole",,"0.6 mole",,),(,,10.8" mole",,):}` water left =11.8-10.8=1 gm `H_(2)SO_(45)=58.8+12=70.8 gm ` |
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