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How many mL of 6.0 M HCl solution will provide 164 g of HCl?

seo-qna
Last updated date: 27th Jul 2024
Total views: 384k
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Answer
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Hint: Here, first we have to calculate the moles of HCl using the formula, Number of moles=$\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{Mass}}}}$. Then, we have to calculate the volume of HCl solution using the formula of molarity.

Complete step by step answer:
Let’s calculate the moles of HCl first. The mass of HCl is given as 164 g . Now, we have to calculate the molar mass of HCl.
Molar mass of HCl=$1 + 35.5 = 36.5\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.
Now, we have to put the values of mass and molar mass of HCl in the formula of moles. So,
The number of moles of HCl=$\dfrac{{164\,{\text{g}}\,{\text{mol}}}}
{{36.5\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}} = 4.49\,{\text{mol}}$
So, the mole of HCl present in the solution is 4.49.
Now, we have to use the formula of molarity to calculate the volume of HCl solution. The formula of molarity is,
Molarity=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}
{{{\text{Volume}}\,{\text{of}}\,{\text{solution}}\left( {\text{L}} \right)}}$
The molality of the HCl solution is given as 6.0 M and the moles of HCl is 4.49. So,
Volume of HCl solution$ = \dfrac{{{\text{Moles of HCl}}}}{{{\text{Molarity}}}}$
$ \Rightarrow $Volume of HCl solution$ = \dfrac{{4.49}}{6} = 0.748\,\,{\text{L}}$
Now, we have to convert the volume in litre to mililitre.
0.748 L=$0.748 \times 1000 = 748\,{\text{mL}}$

Hence, 748 mL of 6.0 M HCl solution will provide 164 g of HCl.

Additional Information:
Molarity is a way of expressing concentration of a solution. Molarity is defined as the number of moles of solute present in one litre solution.

Note: The mole or mol (abbreviation form) is the unit which measures the amount of substance present. A mole is the term that defines the number of carbon (C) atoms in 12 g of pure carbon. After so many years of experiment, it has been proved that a mole of anything contains \[6.022 \times {10^{23}}\] representative particles.