What is the molarity of a $49.0g$ mass of phosphoric acid in a $2.00L$ volume of aqueous solution?

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Answer 1 · 2016-09-18T07:29:54

$"Molarity"$ $=$ $"Moles of solute"/"Volume of solution"$

We get an answer with units $mol*L^-1$ .

$"Moles of phosphoric acid "(H_3PO_4)$ $=$ $(49.0*g)/(97.99*g*mol^-1)$ $=$ $0.5*mol$ .

$"Molarity"$ $=$ $(0.50*mol)/(2000*cm^3xx10^-3cm^3*L^-1)$

$=$ $0.25*mol*L^-1$ with respect to $H_3PO_4$ .

What is the molarity of a 49.0*g49.0*g mass of phosphoric acid in a 2.00*L2.00*L volume of aqueous solution?