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weqwewe [10]
2 years ago
10

4. You have a very concentrated solution (12 M) of potassium chloride (KCI). You need it to be at the least concentration possib

le for the experiment you are
about to conduct. The problem is you forgot to order lab supplies, so you only have 2 L of distilled water left. What would be the final concentration if you
added the two 2 L of distilled water to the 0.5 L of 12 M KCI?
a 3.0 M
b 24 M
C 2.4M
d 48 M
Chemistry
1 answer:
Alenkinab [10]2 years ago
4 0

Answer:

C. 2.4 M

Explanation:

Because you started with 12M solution of KCl, that means that there was a concentration of 12 mols of KCl per Liter of solution. (12mol/1L =12 M). Since there was only 0.5L of solution, there was only 6 mols of KCl because there is only 12 mols per 1 Liter, so half of that volume would have half the amount of solute to keep that true. (6mol/.5L = 12 M). With the new water added, the formula becomes (6mols KCl/ 2.5L of water =2.4 M)

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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

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