Which of the following is/are correct for 17g/L of  $H_2{O}_2$  solution

 $\begin{array}{l}17g/L{H}_2{O}_2=\frac{17}{34}mol{L}^{-1}{H}_2{O}_2\\ Molarityof{H}_2{O}_2=\frac{1}{2}\\ M=\frac{V}{11.2}\Rightarrow V=\frac{1}{2}\times 11.2=5.6\end{array}$
 5.6  Volume  $H_2{O}_2$ means 1mL will give
5.6 mL at 273 K and 1atm
 $\begin{array}{l}{P}_1{V}_1=P_2{V}_2\\ 1\times 5.6=2\times V_2\\ V_2=2.8mL\end{array}$