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Sever21 [200]
3 years ago
9

What is the molality (m) of a solution that contains 76.5 g of KCl dissolved in 85.0 g of water?

Chemistry
1 answer:
ladessa [460]3 years ago
5 0

Answer:

76.5g KCl/74.55 grams per mole Kcl = x

molality= x/.085 kg H2O

Explanation:

well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.

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The balanced equation for the reaction is given below:

<h3>3AgNO₃(aq) + AICI₃(aq) —> Al(NO₃)₃ (aq) + 3AgCl (s) </h3>

Next, we shall determine the mass of aluminum chloride, AICI₃ that reacted and the mass of silver chloride, AgCl produced from the balanced equation. This is illustrated below:

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Mass of AICI₃ from the balanced equation = 1 × 133.34 = 133.34 g

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From the balanced equation above,

133.34 g of AICI₃ reacted to produce 429.96 g of AgCl.

Finally, we shall determine the theoretical yield of AgCl by the reaction of 1.616 g of AICI₃ as follow:

From the balanced equation above,

133.34 g of AICI₃ reacted to produce 429.96 g of AgCl.

Therefore,

1.616 g of AICI₃ will react to produce = \frac{1.616 * 429.96}{133.34} = 5.2109 g of AgCl.

Thus, the theoretical yield of silver chloride, AgCl is 5.2109 g

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