What is the molarity of $50% w/w$ $"50% w/w"$ $H C l$ $HCl$ solution for which $ρ_{solution} = 1.15 \cdot g \cdot c m^{- 3}$ $rho_"solution"=1.15gcm^-3$ ?

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What is the molarity of $50% w/w$ $"50% w/w"$ $H C l$ $HCl$ solution for which $ρ_{solution} = 1.15 \cdot g \cdot c m^{- 3}$ $rho_"solution"=1.15gcm^-3$ ?

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Answer 1 · 2017-07-27T17:18:40

You have not specified a solvent.......and the question has not been written by a chemist......

Explanation:

We assume an aqueous solution of hydrochloric acid, and clearly a $1 \cdot m L$ $1*mL$ volume of the stuff contains $50 % \times 1 \cdot c m^{3} \times 1.15 \cdot g \cdot c m^{- 3} = 0.575 \cdot g$ $50%xx1*cm^3xx1.15*g*cm^-3=0.575*g$ $H C l$ $HCl$ .....

And thus the concentration is....... $\frac{moles of solute}{volume of solution}$ $"moles of solute"/"volume of solution"$

$\frac{\frac{50 % \times 1 \cdot c m^{3} \times 1.15 \cdot g \cdot c m^{- 3}}{36.46 \cdot g \cdot m o l^{- 1}}}{1 \times 10^{- 3} \cdot L} \equiv 15.8 \cdot m o l \cdot L^{- 1}$ $((50%xx1*cm^3xx1.15*g*cm^-3)/(36.46*g*mol^-1))/(1xx10^-3*L)-=15.8*mol*L^-1$ .

The most concentrated hydrochloric acid you can buy is $36 - 38 % (\frac{w}{w})$ $36-38%(w/w)$ ; typically we are supplied conc. $H C l$ $HCl$ as a $10.5 \cdot m o l \cdot L^{- 1}$ $10.5*mol*L^-1$ solution in the laboratory.

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What is the molarity of $50% w/w$ $"50% w/w"$ $H C l$ $HCl$ solution for which $ρ_{solution} = 1.15 \cdot g \cdot c m^{- 3}$ $rho_"solution"=1.15gcm^-3$ ?

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What is the molarity of 50% w/w"50% w/w" HClHCl solution for which ρsolution=1.15⋅g⋅cm−3rho_"solution"=1.15*g*cm^-3?

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What is the molarity of $50% w/w$ $"50% w/w"$ $H C l$ $HCl$ solution for which $ρ_{solution} = 1.15 \cdot g \cdot c m^{- 3}$ $rho_"solution"=1.15gcm^-3$ ?