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) \(2{ H }_{ 2 }{ O }_{ 2 }\left( l \right) \underrightarrow { hv } 2{ H }_{ 2 }O\left( l \right) +{ O }_{ 2 }\left( g \right) \)

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.
\(8LiH+{ Al }_{ 2 }{ Cl }_{ 6 }\longrightarrow 2LiAl{ H }_{ 4 }+6LiCl\)

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

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
\(\frac { 32\times 34 }{ 68 } =16g{ O }_{ 2 }\)