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sergeinik [125]
3 years ago
15

What is the molarity of a solution that contains 1.78 moles of KNO3 dissolved in 2.50L of water?

Chemistry
1 answer:
Alexxandr [17]3 years ago
4 0

Answer:

molarity= 0.712M

Explanation:

From n=CV

n=1.78, C=?, V= 2.5L

1.78=C×2.5

C= 0.712M

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Fluorine (F) would be least likely to form a cation out of potassium, fluorine, chlorine, and nitrogen.

  • A cation is a positively charged atom (or molecule) that has lost electrons (or electrons).
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  • Going down a group increases electropositivity, or the propensity to lose electrons and generate cations. and decreases across a period. In the given examples:
  • Potassium, K is an alkali metal and will lose electrons readily to form a cation.
  • Nitrogen (N), Fluorine (F), and chlorine (Cl) are all nonmetals that prefer to accept electrons and form anions instead. F is the most electronegative i.e. it will gain electrons and form F- rather than F+.

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What is the molarity of a solution prepared by diluting 250 mL of a 40% H2SO4 solution to 1 Liter? The density of the stock solu
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Answer:

1.195 M.

Explanation:

  • We can calculate the concentration of the stock solution using the relation:

<em>M = (10Pd)/(molar mass).</em>

Where, M is the molarity of H₂SO₄.

P is the percent of H₂SO₄ (P = 40%).

d is the density of H₂SO₄ (d = 1.17 g/mL).

molar mass of H₂SO₄ = 98 g/mol.

∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.

  • We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.

<em>∴ (MV) before dilution = (MV) after dilution</em>

M before dilution = 4.78 M, V before dilution = 250 mL.

M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.

∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.

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