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8_murik_8 [283]
3 years ago
11

What is the concentration of a solution with 2.00 moles of solute in 4.00 l of solution?

Chemistry
1 answer:
jarptica [38.1K]3 years ago
6 0

Explanation:

Molarity is defined as the the number of moles present in a solution divided by volume in liters.

Mathematically,    Molarity = \frac{/text{no. of moles}}{volume in liter}

As it is given that number of moles present into the solution are 2.00 moles and volume is 4 liter. Therefore, calculate the molarity of solution as follows.

                Molarity = \frac{/text{no. of moles}}{volume in liter}

                          = \frac{2.00 moles}{4.00 L}

                          = 0.5 M

Thus, we can conclude that the concentration of given solution is 0.5 M.

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Answer:

1) Atomic number = number of proton or electrons

Mass number = number of proton or electron + number of neutron

3 0
4 years ago
Vinegar is a chemical used in cooking, cleaning and other common experiences. A 0.1 M solution of vinegar in water has a [H+] of
Ymorist [56]

Answer: a) pH of a 0.1 M vinegar solution is 2.9

b) It is an acid as pH is less than 7

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.

As vinegar is  a weak acid, its dissociation is represented as;

CH_3COOH\rightleftharpoons H^+CH-3COO^-

   cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.1 M

[H^+]=c\times \alpha

[H^+]=0.1\times \alpha

1.3\times 10^{-3}=0.1\times \alpha

\alpha=0.013

Also pH=-log[H^+]

pH=-log[1.3\times 10^{-3}]=2.9

Thus pH of a 0.1 M vinegar solution is 2.9

As pH is less than 7,  it is an acid.

7 0
4 years ago
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4 years ago
"what is the mole fraction of solute in a 3.87 m aqueous solution?"
sattari [20]

The mole fraction of solute in a 3.87 m aqueous solution is 0.0697

<h3> calculation</h3>

molality = moles of the solute/Kg of the solvent

3.87 m dissolve in 1 Kg of water= 1000g

find the moles of water= mass/molar mass

that is 1000 g/ 18 g/mol= 55.56 moles

mole of solute = 3.87 moles

mole fraction is = moles of solute/moles of solvent

that is 3.87/ 55.56 = 0.0697

5 0
3 years ago
Read 3 more answers
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