0.2 L(or 200 mL) of 5M solution will contain 
\(\frac { 340\times0.2 }{ 2 } =34g{ H }_{ 2 }{ O }_{ 2 }\)
\(\underset { 2\times 34=68g }{ { 2H }_{ 2 }O_{ 2 } } \rightarrow { 2H }_{ 2 }O+\underset { 2\times 16=32g }{ { O }_{ 2 } } \)
\(\because \ 68g \ { H }_{ 2 }O_{ 2 }\) on decomposition will give 32g\(O_{ 2 }\)
\(\therefore \ 34g \ { H }_{ 2 }O_{ 2 }\) on decomposition will give
\(\frac { 32\times 34 }{ 68 } =16g{ O }_{ 2 }\)

Molar mass of \({ H }_{ 2 }{ O }_{ 2 }=34 \ gmol^{ -1 }\)
1L of 5M solution of \({ H }_{ 2 }O_{ 2 }\)will contain 34 x 5g \({ H }_{ 2 }O_{ 2 }\)
2L of 5M solution of \({ H }_{ 2 }O_{ 2 }\)will contain 34 x 5 x 2 = 340g \({ H }_{ 2 }O_{ 2 }\)
Mass of \({ H }_{ 2 }O_{ 2 }\)present in 2L of 5 molar solution = 340g

It is LiH because it has significant covalent character due to the smallest alkali metal Li. LiH is very stable. It is almost unreactive towards oxygen and chlorine.
It reacts with \({ Al }_{ 2 }{ Cl }_{ 6 }\) to form lithium aluminium hydride.
\(8LiH+{ Al }_{ 2 }{ Cl }_{ 6 }\longrightarrow 2LiAl{ H }_{ 4 }+6LiCl\)

Since, a colourless liquid A contains only H and O and decomposes slowly on exposure to light but is stabilised by additions of urea, therefore, liquid A may be hydrogen peroxide.
(ii) \(2{ H }_{ 2 }{ O }_{ 2 }\left( l \right) \underrightarrow { hv } 2{ H }_{ 2 }O\left( l \right) +{ O }_{ 2 }\left( g \right) \)