What is the molar concentration of sulfuric acid is we have an $80% w/w$ $"80% w/w"$ concentration with $ρ_{solution} = 1.73 \cdot g \cdot m L^{- 1}$ $rho_"solution"=1.73gmL^-1$ ?

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What is the molar concentration of sulfuric acid is we have an $80% w/w$ $"80% w/w"$ concentration with $ρ_{solution} = 1.73 \cdot g \cdot m L^{- 1}$ $rho_"solution"=1.73gmL^-1$ ?

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Answer 1 · 2016-03-26T09:54:50

Approx. $14$ $14$ $m o l \cdot L^{- 1}$ $mol*L^-1$

Explanation:

We know that $1$ $1$ $m L$ $mL$ of sulfuric acid has a mass of $1.73$ $1.73$ $g$ $g$ , of which $1.384$ $1.384$ $g$ $g$ is $H_{2} S O_{4}$ $H_2SO_4$ .

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

$(1.384*cancelg)/(98.079*cancelg*mol^-1)$ $xx$ $1/(1*cancel(mL)xx(10^-3L*cancel(mL^-1))$ $=$ $?? mol*L^-1$

$"98% sulfuric acid (w/w)"$ has a density of $1.83*g*mL^-1$ . What is the molarity of this solution.

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What is the molar concentration of sulfuric acid is we have an $80% w/w$ $"80% w/w"$ concentration with $ρ_{solution} = 1.73 \cdot g \cdot m L^{- 1}$ $rho_"solution"=1.73gmL^-1$ ?

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What is the molar concentration of sulfuric acid is we have an "80% w/w"80% w/w"80% w/w" concentration with rho_"solution"=1.73*g*mL^-1ρsolution=1.73⋅g⋅mL−1rho_"solution"=1.73*g*mL^-1?

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What is the molar concentration of sulfuric acid is we have an $80% w/w$ $"80% w/w"$ concentration with $ρ_{solution} = 1.73 \cdot g \cdot m L^{- 1}$ $rho_"solution"=1.73gmL^-1$ ?