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Leona [35]
4 years ago
13

How would you prepare 3.5 L of a 0.9M solution of KCl?

Chemistry
1 answer:
Fudgin [204]4 years ago
8 0
Calculate the mass of the solute <span>in the solution :

Molar mass KCl = </span><span>74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g 

</span><span>To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm</span>³<span> and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!</span>
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<h3>Explanation:</h3>

<u>We are given;</u>

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But, K = °C + 273

          = -15°C + 273

          = 258 K

  • Initial volume, V₁ =3.48 L
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                                   = 36 °C + 273

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P₁V₁/T₁ = P₂V₂/T₂

  • We can calculate the new volume, V₂;
  • Rearranging the formula;

V₂ =(P₁V₁T₂) ÷ (T₁P₂)

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