Question

find molarity of h + ions if 49 gram H2 S o4 is dissolved in 250 ml solution​

Answer 1

Answer:

The molarity of the $H^+$ is 4 mol/L.

Explanation:

$H_2SO_4\rightarrow 2H^++SO_{4}^{2-}$

Molarity of sulfuric acid

$M=\frac{49 g}{98 g/mol\times 0.250 L}=2 mol L$

$[H_2SO_4]=2 mol/L$

1  mole of sulfuric acid gives 2 moles of $H^+$ ions.

The concentration of $H^+$ ions is:

$2\times [H_2SO_4]=2\times 2 mol/L=4 mol/L$

The molarity of the $H^+$ is 4 mol/L.

Answer 2

The molarity of hydrogen ions in the given solution is 4 M

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of sulfuric acid = 49 grams

Molar mass of sulfuric acid = 98 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfuric acid}=\frac{49\times 1000}{98\times 250}\\\\\text{Molarity of sulfuric acid}=2M$

1 mole of sulfuric acid produces 2 moles of hydrogen ions 1 mole of sulfate ions

So, molarity of $H^+$ ions = (2 × 2) = 4 M

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