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Pavlova-9 [17]
3 years ago
15

The molar concentration (M) of a solution prepared by dissolving 0.2362g of Cr(NO3)3 in a 50-mL volumetric flask is 0.01985M, wh

ere the molecular weight for Cr(NO3)3 = 238.01g/mol.
a. Suppose you want to prepare another solution containing chromium nitrate that is 25 times LESS concentrated than the one prepared above. Given a choice of 10-mL and 5-mL pipets and 50-mL and 100-mL volumetric flasks, explain how you would proceed in preparing the new diluted solution. In addition, calculate the concentration for the new diluted solution. SHOW ALL WORK. Your final value should have the correct unit and number of significant figures. Hint: You will most likely need two dilution steps in order to obtain the desired concentration. Note: You may NOT reuse the same pipet or combine different pipets within the same dilution step. You may reuse the pipet and/or volumetric flask in the different dilution step.
Chemistry
1 answer:
kati45 [8]3 years ago
3 0

Answer:

Use a 10 mL pipet and a 50 mL volumetric flask, <em>twice</em>

Explanation:

Given those conditions, in order to have a solution that is 25 times less concentrated that the one previously prepared, we would need two dillution steps, as suggested by the problem.

If we take an aliquot of 10 mL and dilute it to a final volume of 50 mL, we prepare a solution that is 5 times less concentrated. And if we then take again an aliquot of 10 mL of that dilluted solution and dilute it to a final volume of 50 mL, we would have a solution 25 (5 times 5) times less concentrated than the original one.

In order to verify, we can calculate the resulting concentration and compare it with the original one:

0.01985M*\frac{10mL}{50mL} *\frac{10mL}{50mL}=0.000794M

Thus we can see that the obtained concentration is in fact 25 times lower than the original concentration.

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When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
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Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

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Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

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