We’re being asked to calculate the **molarity (M)** of a solution of H_{2}SO_{4}. Recall that ** molarity** is the ratio of the moles of solute and the volume of solution (in liters). In other words:

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{\left(}}{\mathbf{M}}{\mathbf{\right)}}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{}\mathbf{mole}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

We first need to determine the **number of moles of ****H _{2}SO**

We’re given the mass of H_{2}SO_{4}. We can use the molar mass of H_{2}SO_{4} to find the moles.

**The molar mass of H _{2}SO_{4} is:**

H_{2}SO_{4} 2 H x 1.008 g/mol H = 2.016 g/mol

1 S x 32.07 g/mol S = 32.07 g/mol

4 O x 16.00 g/mol O = 64.00 g/mol

Molar mass = *98.086g/ 1 mol H _{2}SO_{4}*

Let's consider a solution made by dissolving 2.355 g of sulfuric acid, H _{2}SO_{4}, in water. The total volume of the solution is 50.0 mL. What is the molar concentration of sulfuric acid, [H_{2}SO_{4}]?

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