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yawa3891 [41]
3 years ago
12

An aqueous solution is found to be 3.64% potassium nitrate by mass, how many grams of solution should I evaporate the water from

to get it 18.7 g of potassium nitrate
Chemistry
1 answer:
Sidana [21]3 years ago
8 0

Answer:

513.74 g of solution

Explanation:

% Mass grams are defined as the <em>grams that are dissolved in salt</em> (in this case, it would be <em>potassium nitrate</em>) <em>dissolved every 100 g of the solution</em>. Having this information, you can calculate the amount of solution that has dissolved 18.7 g of potassium nitrate, which is what we want to obtain.

The relationship is:

3.64 g of potassium nitrate _____ 100 g solution

18.7 g of potassium nitrate _____ X = 513.74 g of solution

Calculation: 18.7g x 100g / 3.64g = 513.74 g of solution

So, <em>I need 513.74 g of solution to get 18.7g of potassium nitrate by evaporating it</em>.

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1. Find the masses of the following amounts.
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The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

<h3>What is Mole ?</h3>

A mole is a very important unit of measurement that chemists use.

A mole of something means you have 6.023 x 10 ²³ of that thing.

  • For 2.15 mol of hydrogen sulphide (H₂S) :

1 mole hydrogen sulphide (H₂S) = 34.08088 grams

Therefore,

2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol

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  • For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;

1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams

Therefore,

3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol

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Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of  3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm

Learn more about mole here ;

brainly.com/question/21323029

#SPJ1

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Explanation:

Hope u got ur answer

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