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algol [13]
3 years ago
12

Which aqueous solution of KI freezes at the lowest temperature?

Chemistry
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

b. 2 mol of KI in 500. g of water

Explanation:

We have to apply the colligative property of freezing point depression.

The formula is: ΔT = Kf . m . i

As the (Kf . m . i) is higher, then the freezing temperature will be lower.

i refers to the Van't Hoff factor (number of ions dissolved in the solution)

KI → K⁺ + I⁻    (i =2)

Kf is constant so, we have to search for the highest m (molality)

Molality means the moles of solute in 1kg of solvent.

The highest m is option b → 2 mol of KI / 0.5 kg = 4 mol/kg

a. 1 mol of KI / 0.5 kg = 2 mol/kg

c. 1 mol of KI / 1kg = 1 mol/kg

d. 2 mol of KI / 1kg = 2 mol/kg

1000 g = 1kg. In order to determine molality we need to convert the mass (g) of solvent to kg

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What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen
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3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

C_3H_8(g) + 5 O_2(g)→ 3 CO_2(g) + 4 H_2O(g)

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99  x 0.08206 x 623) ÷ 0.96

v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

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