Question

Find the volume strength of 1.6 N $H_{2}O_{2}$ solution.

Answer 1

$Strength\quad =\quad Normality\quad \times \quad Equivalent\quad weight$

$Equivalent\quad weight\quad of\quad { H }_{ 2 }{ O }_{ 2 }\quad =\quad 17\\ Strength\quad of\quad 1.6\quad N\quad { H }_{ 2 }{ O }_{ 2 }\quad solution\quad =\quad 1.6\quad \times \quad 17\quad g{ L }^{ -1 }$

$Now\quad 68g\quad of\quad { H }_{ 2 }{ O }_{ 2 }\quad gives\quad 22400\quad ml\quad of\quad { O }_{ 2 }\quad at\quad NTP/STP$

$1.6\times 17g\quad of\quad { H }_{ 2 }{ O }_{ 2 }\quad will\quad give\quad$ $=\quad \frac { 22400 }{ 68 } \quad \times \quad 16\quad \times \quad 17\quad =\quad 8960\quad ml\quad of\quad { O }_{ 2 }$

$Strength\quad =\quad Normality\quad \times \quad Equivalent\quad weight$

$Equivalent\quad weight\quad of\quad { H }_{ 2 }{ O }_{ 2 } = 17$

$Strength\quad of\quad 1.6\quad N\quad { H }_{ 2 }{ O }_{ 2 }\quad solution\quad =\quad 1.6\quad \times \quad 17\quad g{ L }^{ -1 }$

$But\quad 1.6\quad \times \quad 17g\quad of\quad { O }_{ 2 }\quad are\quad present\quad in\quad 1000\quad ml\quad of\quad { O }_{ 2 }\quad solution$

Hence,$\quad 1000\quad ml\quad of\quad { H }_{ 2 }{ O }_{ 2 }\quad solution\quad gives\quad 8960\quad ml\quad of\quad { O }_{ 2 }\quad at\quad STP$

Hence,$\quad the\quad voluem\quad strength\quad of\quad 1.6N\quad { H }_{ 2 }{ O }_{ 2 }\quad solution\quad is\quad =\quad 8.96\quad volume$