Question
  1. Commercially available concentrated hydrochloric acid contains $45\%$ HCl by mass.
  1. What is the molarity of this solution? The density is $1.19g$ mL.
  2. What volume of cone. HCl is required to make $1.00L$ of $0.24M$ HCl?
  1. Write the balanced chemical equations for the following:
  1. $KMnO_4 + NH3 \rightarrow MnO_2 +KOH+ H_2O + N_2$
  2. $HNO_3 + P_4\rightarrow NO_2 + H_3PO_4 + H_2O$

Answer

  1.  
  1. 45% solution, means 45g of HCl in 100g of solution.
Then, Mass of the solution = 100g
Volume of the solution $=\frac{100\text{g}}{\text{Density}}=\frac{100\text{g}}{1.19\text{g/ mL}}$
$=840\text{mL}=\frac{84.0}{1000}\text{L}$
Molar mass of HCl = $36.5g mol^{-1}$
No. of moles of HCl dissolved $=\frac{45}{36.5}\text{mol}=1.23\text{mol}$
$\therefore$ Molarity of HCl solution $=\frac{1.23}{\Big(\frac{84}{1000}\Big)\text{L}}=14.64\text{mol L}^{-1}$
  1. Molarity or conc. HCl sample = 14.64mol/ L
Molarity of HCl solution to be prepared = 0.24mol/ L
Volume of HCl solution to be prepared = 1.00L = 1000mL
Then, using the molarity equation, MV - MV,
$\text{V}_{\text{HCl}}=\frac{0.24}{14.64}=16.40$
Thus, to obtain 1.0L of 0.24M HCl, one should dissolve 16.40mL of conc. HCl to make up the volume to 1.0L.
  1.  
  1. $2KMnO_4 + 2NH_3 \rightarrow 2MnO_4 + 2KOH + 2H_2O + N_2$
  2. $2OHNO_3 + P_4\rightarrow 2NO_2 + 4H_3PO_4 + 4H_2O$

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