A $1.0\ \mathrm{g}$ sample containing $\ce{BaCl2*2H2O}$ was dissolved and an excess of $\ce{K2CrO4}$ solution added. After a suitable period, the $\ce{BaCrO4}$ was filtered, washed and redissolved in $\ce{HCl}$ to convert $\ce{CrO4^2-}$ to $\ce{Cr2O7^2-}$. An excess of $\ce{KI}$ was added, and the liberated iodine was titrated with $84.7\ \mathrm{mL}$ of $0.137\ \mathrm{M}$ sodium thiosulphate. Calculate the percent purity of $\ce{BaCl2*2H2O}$.
I first found the number of mili-equivalents of sodium thiosulphate and equated that to the number of moles of pure $\ce{BaCl2*2H2O}$ times the n-factor i.e. $2$.
From this equation I get the number of moles of pure $\ce{BaCl2*2H2O}$ and then multiply this by its molecular mass to get the mass of pure $\ce{BaCl2*2H2O}$. But it comes greater than $1\ \mathrm{g}$. How to solve this?