Question

A commercially available sample of $H_{2}SO_{4}$ is 15% $H_{2}SO_{4}$ by weight (density = 1.10g/ml). Calculate molarity of the solution.

Answer 1

From the given density = 1.10 g/mol

$Molecular\quad weight\quad of\quad { H }_{ 2 }{ SO }_{ 4 }\quad =\quad \left( 2\quad \times \quad 1 \right) \quad +\quad 32\quad +\quad 64\quad =\quad 98\quad g\quad mo{ l }^{ -1 }M\quad \\\\=\quad \frac { { W }_{ B } }{ { M }_{ B }\quad \times \quad { V }_{ L } } M\quad \\ \\=\quad \frac { 15\quad g\quad \times \quad 1000 }{ 98\quad g/mol\quad \times \quad 115\quad g\quad \times \quad 1.10\quad g/c{ m }^{ 3 } } \\ \\=\quad \frac { 15000 }{ 12397 } molar\\ \\=\quad 1.209\quad molar$

$Molarity\quad =\quad \frac { { W }_{ B }\quad \times \quad 1000 }{ { W }_{ B }\quad \times \quad 100\quad g } \\ \\=\quad \frac { 15\quad g\quad \times \quad 10\quad}{98\quad g\quad mo{ l }^{ -1 } } \\\\=\quad \frac { 150\quad}{ 98 } \\ \\Molarity\quad =\quad 1.53\quad molar$