## Chemistry 12th Edition

We know that: $Molar\space mass \space of Ba=137.3$ $Molar\space mass\space of BaSO_4=233.6$ $Mass\space of\space Brium in BaSO_4=(\frac{0.4105g\space BaSO_4\times 137.3g\space Ba}{233.4g\space BaSO_4})=0.2415g\space Ba$ $\%Ba\space by \space mass=(\frac{experimental}{theoretical }\times 100)=\frac{0.2415}{0.676g}\times 100\%=35.7\%$