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

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

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Answer 1 · 2016-09-18T07:29:54

$Molarity$ $"Molarity"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$

Explanation:

We get an answer with units $m o l \cdot L^{- 1}$ $mol*L^-1$ .

$Moles of phosphoric acid (H_{3} P O_{4})$ $"Moles of phosphoric acid "(H_3PO_4)$ $=$ $=$ $\frac{49.0 \cdot g}{97.99 \cdot g \cdot m o l^{- 1}}$ $(49.0*g)/(97.99*g*mol^-1)$ $=$ $=$ $0.5 \cdot m o l$ $0.5*mol$ .

$Molarity$ $"Molarity"$ $=$ $=$ $\frac{0.50 \cdot m o l}{2000 \cdot c m^{3} \times 10^{- 3} c m^{3} \cdot L^{- 1}}$ $(0.50*mol)/(2000*cm^3xx10^-3cm^3*L^-1)$

$=$ $=$ $0.25 \cdot m o l \cdot L^{- 1}$ $0.25*mol*L^-1$ with respect to $H_{3} P O_{4}$ $H_3PO_4$ .

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

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Explanation:

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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?

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Explanation:

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