A 3.0 L solution contains 73.5 g of $H_{2} S O_{4}$ $H_2SO_4$ . What is the molar concentration of the solution?

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A 3.0 L solution contains 73.5 g of $H_{2} S O_{4}$ $H_2SO_4$ . What is the molar concentration of the solution?

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Answer 1 · 2017-06-27T16:43:39

Approx. $0.250 \cdot m o l \cdot L^{- 1}$ $0.250*mol*L^-1$ .

Explanation:

$Concentration = \frac{\frac{73.5 \cdot g}{98.08 \cdot g \cdot m o l^{- 1}}}{3.0 \cdot L} = 0.250 \cdot m o l \cdot L^{- 1}$ $"Concentration"=((73.5*g)/(98.08*g*mol^-1))/(3.0*L)=0.250*mol*L^-1$

The concentration is quoted with respect to $H_{2} S O_{4} (a q)$ $H_2SO_4(aq)$ . What is the concentration with respect to $H_{3} O^{+}$ $H_3O^+$ , and why do I make this distinction?

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A 3.0 L solution contains 73.5 g of $H_{2} S O_{4}$ $H_2SO_4$ . What is the molar concentration of the solution?

1 Answer

Explanation:

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A 3.0 L solution contains 73.5 g of $H_{2} S O_{4}$ $H_2SO_4$ . What is the molar concentration of the solution?