Dnyanesh Vernekar - Chemistry
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The balanced equations for the precipitation of Cr^{3+} and Mg^{2+} using NaF can be written as below
Cr^{3+} + 3NaF^{ }$\to $CrF_{3 }+ 3Na^{+}
Mg^{2+ }+ 2NaF $\to $MgF_{2 }+ 2Na^{+}
The molecular weight (or molar mass) of CrF_{3 }= (52+3x19)= 109.0 g
The molecular weight (or molar mass) of MgF_{2} = (24.3 +2x19) = 62.3 g
Molarity (M) of the solution is defined as the number of moles of the solute present (dissolved) in 1 liter of the solution.
The concentration of NaF is given as 1.51 M.
To get mol of F
$\frac{molofF}{1L}=1.51M\phantom{\rule{0ex}{0ex}}molofF=1.51Mx1L=1.51$
Thus 1.51 moles of F^{-} ions are used to precipitate.
Also we know that,
mass of CrF_{3} + mass of MgF_{2} = total mass of precipitate
if x = mass of CrF_{3} and y =mass of MgF_{2}
then x + y= 50.1 ....... equation 1
From the balanced equation given in Step 1
we require 3 mol of F to get 1 mol of CrF_{3 }and 2 mol of F to get 1 mol of MgF_{2}
Hence, we can represent this in equation form as
3 (mol of CrF_{3}) + 2 (mol of MgF_{2) }= Mol of F
By definition
Mol = weight/molecular weight
Hence,
$\frac{3\times mass\left(CrF3\right)}{Molecularweight\left(CrF3\right)}+\frac{2\times mass\left(MgF2\right)}{Molecularweight\left(MgF2\right)}=MolofF$
substituting the values we get
$\frac{3x}{109}+\frac{2y}{62.3}=1.5$ ............... equation 2
simplifying Equation 2, we get
$2y=62.3\times (1.5-\frac{3x}{109})\phantom{\rule{0ex}{0ex}}2y=62.3\times \left(\frac{163.5-3x}{109}\right)\phantom{\rule{0ex}{0ex}}y=0.285\times (163.5-3x)\phantom{\rule{0ex}{0ex}}y=46.76-0.856x$
substituting this value of y in equation 1, we get
$x+46.76-0.857x=50.1\phantom{\rule{0ex}{0ex}}0.144x=50.1-46.76\phantom{\rule{0ex}{0ex}}0.144x=3.34\phantom{\rule{0ex}{0ex}}x=23.19g$
Thus 23.19g of CrF_{3} is precipitated.
Mol = weight/molecular weight
hence, mol of Cr originally present = 23.19/109.0
Mol of Cr = 0.213 moles
number of moles = weight/ molecular weight
molecular weight of Cr is 52
Hence mass of Cr^{3+} in the original solution = 0.213 x52 = 11.08g
The mass of Cr^{3+} is 11.08g in the original solution