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puteri [66]
2 years ago
7

What is the concentration of a KCl solution made by diluting 75.0 mL of a 0.200 M solution to a final volume of 100. mL?

Chemistry
1 answer:
ss7ja [257]2 years ago
6 0

Answer:

0.15 M KCl

Explanation:

(Step 1)

Convert the original volume from mL to L. Calculate the moles of the original solution.

75.0 mL / 1,000 = 0.0750 L

Molarity = moles / volume (L)

0.200 M = moles / 0.0750 L

0.0150 = moles

(Step 2)

Convert the final volume from mL to L. Calculate the final molarity.

100. mL / 1,000 = 0.100 L

Molarity = moles / volume (L)

Molarity = 0.0150 moles / 0.100 L

Molarity = 0.15 M

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How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass
sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

Learn more about precipitation here:

brainly.com/question/13859041

#SPJ4

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