Question

Calculate the molarity of 152.0g of H2SO4 in 1.070L of solution

Answer 1

1.448 Mol/L

Step-by-step explanation

We need to first calculate the mass of H2SO4 in 1.070 L as follows

`1\text{ L solution }*\frac{152.0\text{ g H2SO4}}{1.070\text{ L solution}}=142.06\text{ g of H2SO4}`

Now the molarity will be calculated as follows

Note: Molar mass of H2SO4 = 98.08 g/mol

H2SO4 = (2 x 1.00784) +32.065 + (4 x 15.999) = 98.08 g/mol

`\begin{gathered} Molarity=142.06\text{ g}*\frac{1\text{ mole H2SO4}}{98.08\text{ g/mole}} \\ \text{Molarity = 1.448 Mol L}^(-1) \end{gathered}`