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Maurinko [17]
3 years ago
7

The molarity of a solution of 5.0 g of kcl in 100. ml of solution is ________.

Chemistry
2 answers:
Sliva [168]3 years ago
8 0

<u>Answer:</u> The molarity of solution is 0.67 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (KCl) = 5.0 g

Molar mass of potassium chloride = 74.55 g/mol

Volume of solution = 100 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{5g\times 1000}{74.55g/mol\times 100mL}\\\\\text{Molarity of solution}=0.67M

Hence, the molarity of solution is 0.67 M

sdas [7]3 years ago
7 0
Molarity is defined as the number of moles of solute in 1 L of solution 
mass fo KCl in the solution is - 5.0 g
number of moles of KCl - 5.0 g/ 74.5 g/mol  = 0.067 mol
number of moles of KCl in 100 mL - 0.067 mol
therefore number of KCl moles in 1 L - 0.067 / 100 mL x 1000 mL = 0.67 M
molarity of KCl is 0.67 M
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A student made a copper bracelet by hammering a small copper bar into the desired shape. The bracelet has a mass of 30.1 grams a
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69. The student's skin was obviously warmer than the initial temperature of the bracelet. According to the second law of thermodynamics, heat transfers from a higher temperature to a lower temperature hence heat transfers from the student's skin to the bracelet warming up the bracelet. 

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The atmosphere of the planet Mars is 95.3% carbon dioxide, 2.6% nitrogen, 1.9% argon, with the remaining fraction being traces o
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Explanation:

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Hope this Helps!!!

3 0
3 years ago
3. In a typical titration experiment a student titrates a 5.00 mL sample of formic acid (HCOOH), a monoprotic organic acid, with
ira [324]

Answer:

2.893 x 10⁻³ mol NaOH

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Explanation:

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At the endpoint in the titration, the amount of base added is just enough to react with all the formic acid present. So first we will calculate the moles of base added and use the molar ratio from the reaction equation to find the moles of formic acid that must have been present. Then we can find the concentration of formic acid.

The moles of base added is calculated as follows:

n = CV = (0.1088 mol/L)(26.59 mL) = 2.892992 mmol NaOH

Extra significant figures are kept to avoid round-off errors.

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(2.892992 mmol NaOH)(1 HCOOH/1 NaOH) = 2.892992 mmol HCOOH

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7 0
3 years ago
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Answer:

a)<em> Balanced  equation:</em>

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b)<em>Reason:</em>

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Explanation:

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