A 2.0g sample containing MnO₂ is treated with HCl liberating Cl₂. The Cl₂ gas is passed into a solution of Kl and 60.0mL of 0.1M Na₂S₂O₃ is required to titrate the liberated iodine. The percentage of MnO₂ in the sample is_______. (Nearest integer)

[Atomic masses (in u) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32]

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$  

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$               

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}*1000=M*n-factor*V$                              

$\frac{w}{87/2}*1000=0.1*1*60$                              

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample =   $\frac{0.261}{2}*100$

= 13.05%