If 0.2 mol of sulphuric acid is poured into | vedclass

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(B) We know that concentration is dependent on the number of moles of solute and the volume of solution in $L$ and it is given by:Molarity $= \frac{

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$0.8 \ M$

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(B) We know that concentration is dependent on the number of moles of solute and the volume of solution in $L$ and it is given by:Molarity $= \frac{	ext{number of moles of solute}}{	ext{volume of solution in } L}$Here,$n = 0.2 \ mol$,$V = 250 \ mL = 0.25 \ L$Therefore,Molarity $= \frac{0.2}{0.25} = 0.8 \ M$Hence,the concentration of the solution containing $0.2 \ mol$ of sulphuric acid in $250 \ mL$ of water will be $0.8 \ M$.

If $0.2 \ mol$ of sulphuric acid is poured into $250 \ mL$ of water,calculate the concentration of the solution?

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