
How do you prepare a 1 M $HCl$ solution which has a specific gravity of 1.18? The $HCl$solution has 35% purity. Calculate the volume required to prepare the solution.
Answer
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Hint 1 M $HCl$ means there is one mole of $HCl$ in 1 L of water because M is the molarity of the solution. First, calculate the moles of $HCl$ in the solution then find the mass of those moles of $HCl$. 35% purity of the $HCl$ solution means 35 g of $HCl$ is present.
Complete step by step answer:
1 M $HCl$ means there is one mole of $HCl$ in 1 L of water because M is the molarity of the solution. So, for making a solution of 35% purity then we have to first find the number of moles of $HCl$ and since the volume is not specified let us assume a solution is made in 1 L of 1 mol/L $HCl$.
So, the moles of $HCl$will be:
$1L\text{ x }\dfrac{1\text{ }mol}{1\text{ }L}\text{ = 1 mol}$
Now, we have to find the mass of one mole of $HCl$: the mass of hydrogen is 1 and the mass of chlorine is 35.5. So, the mass will be:
$1+35.5=36.5g$
35% purity of the $HCl$ solution means 35 g of $HCl$is present. So, the mass of the solution to make the 35% purity will be:
$36.5\text{ g x }\dfrac{100\text{ g}}{35\text{ g}}\text{ = 104}\text{.2 g}$
So, the volume required to make the 1 M $HCl$of solution having the specific gravity of 1.18 will be:
$\dfrac{104.2}{1.18}\text{ = 88 mL}$
Therefore, the volume required to make the solution is 88 mL.
Note: When we take 88 ml of the concentrated solution, then it must be made up to 1 L of the solution because the molarity of the solution is only measured in the L of the solution.
Complete step by step answer:
1 M $HCl$ means there is one mole of $HCl$ in 1 L of water because M is the molarity of the solution. So, for making a solution of 35% purity then we have to first find the number of moles of $HCl$ and since the volume is not specified let us assume a solution is made in 1 L of 1 mol/L $HCl$.
So, the moles of $HCl$will be:
$1L\text{ x }\dfrac{1\text{ }mol}{1\text{ }L}\text{ = 1 mol}$
Now, we have to find the mass of one mole of $HCl$: the mass of hydrogen is 1 and the mass of chlorine is 35.5. So, the mass will be:
$1+35.5=36.5g$
35% purity of the $HCl$ solution means 35 g of $HCl$is present. So, the mass of the solution to make the 35% purity will be:
$36.5\text{ g x }\dfrac{100\text{ g}}{35\text{ g}}\text{ = 104}\text{.2 g}$
So, the volume required to make the 1 M $HCl$of solution having the specific gravity of 1.18 will be:
$\dfrac{104.2}{1.18}\text{ = 88 mL}$
Therefore, the volume required to make the solution is 88 mL.
Note: When we take 88 ml of the concentrated solution, then it must be made up to 1 L of the solution because the molarity of the solution is only measured in the L of the solution.
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