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trasher [3.6K]
3 years ago
9

Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s

oluble in water. Which produces the largest number of dissolved particles per mole of dissolved solute
Chemistry
1 answer:
slava [35]3 years ago
3 0

Answer:

Potassium carbonate (K₂CO₃)

Explanation:

The compounds dissociate into ions in water, as follows:

K₂CO₃ → 2 K⁺ + CO₃⁻    ⇒ 3 dissolved particles per mole

NaI → Na⁺ + I⁻    ⇒ 2 dissolved particles per mole

KBr → K⁺ + Br⁻   ⇒ 2 dissolved particles per mole

CH₃OH → CH₃O⁻ + H⁺  ⇒ 2 dissolved particles per mole

NH₄Cl → NH₄⁺ + Cl⁻   ⇒ 2 dissolved particles per mole

Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).

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If the solubility of sodium chloride is 36 grams per 100 grams of water, which of the following solutions would be considered un
Murrr4er [49]

Answer: If the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

Explanation:

A solution which contains the maximum amount of solute is called a saturated solution. Whereas a solution in which more amount of solute is able to dissolve is called an unsaturated solution.

Now, the number of moles present in 36 g of NaCl (molar mass = 58.4 g/mol) is as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{36 g}{58.4 g/mol}\\= 0.616 mol

This shows that solubility of sodium chloride is 36 grams per 100 grams of water means a maximum of 0.616 mol of NaCl will dissolve in 100 mL of water.

So, a solution in which number of moles of NaCl are less than 0.616 mol per 100 mL then the solution formed will be an unsaturated solution.

  • As 5.8 moles of NaCl dissolved in 1 L (or 1000 mL) of water. So, moles present in 100 mL are calculated as follows.

Moles = \frac{5.8 mol}{1000 mL} \times 100 mL\\= 0.58 mol

  • Moles present in 100 mL of water for 3.25 moles of NaCl dissolved in 500 ml in water are as follows.

Moles = \frac{3.25 mol}{500 mL} \times 100 mL\\0.65 mol

  • Moles present in 100 mL of water for 1.85 moles of NaCl dissolved in 300 ml of water are as follows.

Moles = \frac{1.85 mol}{300 mL} \times 100 mL\\= 0.616 mol

Thus, we can conclude that if the solubility of sodium chloride is 36 grams per 100 grams of water then 5.8 moles of NaCl dissolved in 1 L of water solution would be considered unsaturated.

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What mass of FeSO4^2- x 6H20 (Molar Mass=260g/mol) is required to produce 500 mL of a .10M iron (II) sulfate solution.
lukranit [14]

Answer:

The correct option is: B. 13g

Explanation:

Given: Molar mass of iron (II) sulfate: m = 260g/mol,

Molarity of iron (II) sulfate solution: M =  0.1 M,

Volume of iron (II) sulfate solution: V = 500 mL = 500 × 10⁻³ = 0.5 L           (∵ 1L = 1000mL)

Mass of iron (II) sulfate taken: w = ? g

<em>Molarity</em>: M = \frac{n}{V (L)} = \frac{w}{m\times V(L)}

Here, n- total number of moles of solute, w - given mass of solute, m- molar mass of solute, V- total volume of solution in L

∴ <em>Molarity of iron (II) sulfate solution:</em>  M = \frac{w}{m\times V(L)}

⇒  w = M\times m\times V(L)

⇒  w = (0.1 M)\times (260g/mol)\times (0.5L)

⇒  <em>mass of iron (II) sulfate taken:</em> w = 13 g

<u>Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.</u>

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What occurs during the death of a large star to make all heavy elements?
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