A 3.0 L solution contains 73.5 g of H2SO4. Calculate the molar concentration of the solution.
Molar mass H2SO4 = 98 g/mol
number of moles:
73,5 / 98 => 0.75 moles
Volume = 3.0 L
M = n / V
M = 0.75 / 3.0
M = 0.25 mol/L
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Answer: 0.25M
Explanation:
Molarity : It is defined as the number of moles of solute present in one liter of solution.
\(Molarity=\frac{\text{given mass of compound}}{\text{molar mass of compound}\times \text{volume of solution is L}}\)
Given mass of \(H_2SO_4\) = 73.5 g
Molar mass of \(H_2SO_4\) = 98g/mol
Volume of solution = 3.0L
\(Molarity=\frac{73.5g}{98g/mol\times 3L}=0.25M\)