9.8 g of H_2SO_4 is present in 2 L of a sol | vedclass

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(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$. Given mass of $H_2SO_4 = 9.8 \ g$. Volume of solution = $2 \ L$. Molarity $(M)$ is defined as the num

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$0.05$

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(B) The molar mass of $H_2SO_4$ is $98 \ g/mol$. Given mass of $H_2SO_4 = 9.8 \ g$. Volume of solution = $2 \ L$. Molarity $(M)$ is defined as the number of moles of solute per liter of solution. Number of moles of $H_2SO_4 = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{9.8 \ g}{98 \ g/mol} = 0.1 \ mol$. Molarity $(M) = \frac{	ext{Number of moles of solute}}{	ext{Volume of solution in liters}} = \frac{0.1 \ mol}{2 \ L} = 0.05 \ M$.

$9.8 \ g$ of $H_2SO_4$ is present in $2 \ L$ of a solution. The molarity of the solution is $....... \ M$

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