Question

Concentrated aqueous sulphuric acid is 98% $H_{2}SO_{4}$ by mass and has a density of 1.84 g $mL^{-1}$. What volume of the concentrated acid is required to make 5.0L of 0.50 M $H_{2}SO_{4}$ solution? ( Mol. weight of sulphuric acid=98)

Answer 1

From the given,

Density of ${ H }_{ 2 }{ SO }_{ 4 }\quad =\quad 1.84\quad g/ml$

Molecular weight of ${ H }_{ 2 }{ SO }_{ 4 }\quad =\quad 98\quad g$

$0.50\quad M\quad { H }_{ 2 }{ SO }_{ 4 }\quad =\quad 0.5\quad \times \quad 98\quad =\quad 49g\quad of\quad { H }_{ 2 }{ SO }_{ 4 }$

To prepare 5L of 0.5 M ${ H }_{ 2 }{ SO }_{ 4 }$, Weight of ${ H }_{ 2 }{ SO }_{ 4 }$ required is $0.5\quad =\quad \frac { Weight\quad of\quad { H }_{ 2 }{ SO }_{ 4 }\quad \times \quad 100 }{ gram\quad molecular\quad weight\quad of\quad { H }_{ 2 }{ SO }_{ 4 }\quad \times \quad Volume\quad of\quad solution\quad in\quad L }$

$Weight\quad of\quad { H }_{ 2 }{ SO }_{ 4 }\quad =\quad \frac { 0.5\times \quad 98\times \quad 5000 }{ 1000 } \quad =\quad 0.245\quad g$

$Weight\quad of\quad 98percent\quad H_{2}SO_{4} =$ $\frac{245\times\ 100}{10\times 98} = 25$

$Volume\quad of\quad { H }_{ 2 }{ SO }_{ 4 }\quad =\quad \frac { Mass\quad  }{ Density } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 25 }{ 1.84 } \quad =\quad 13.5\quad ml$