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# A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M NaF solution is required to cause the complete preci...

Dnyanesh Vernekar - Chemistry

If the question is not under your name, decline under insufficient subject knowledge. Do not Reroute to other test subjects Transcribed: A solution contains Cr3+ and Mg2+. The addition of 1.00 L of 1.51 M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MGF2(s). The total mass of the precipitate is 50.1 g. Find the mass of Cr3+ in the original solution.

The balanced equations for the precipitation of Cr3+ and Mg2+ using NaF can be written as below

Cr3+ + 3NaF $\to$CrF+ 3Na+

Mg2+ + 2NaF $\to$MgF+ 2Na+

The molecular weight (or molar mass) of CrF= (52+3x19)= 109.0 g

The molecular weight (or molar mass) of MgF2 = (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

Thus 1.51 moles of F- ions are used to precipitate.

Also we know that,

mass of CrF3 + mass of MgF2 = total mass of precipitate

if x = mass of CrF3 and y =mass of MgF2

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 CrF3 and 2 mol of F to get 1 mol of MgF2

Hence, we can represent this in equation form as

3 (mol of CrF3) + 2 (mol of MgF2) = Mol of F

By definition

Mol = weight/molecular weight

Hence,

substituting the values we get

$\frac{3x}{109}+\frac{2y}{62.3}=1.5$      ............... equation  2

simplifying  Equation 2, we get

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 CrF3 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 Cr3+ in the original solution = 0.213 x52 = 11.08g

The mass of Cr3+ is 11.08g in the original solution

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