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Andreas93 [3]
2 years ago
11

Potassium fluoride (KF, a salt, has a molecular weight of 58.10 grams. How many grams would be needed to make 1.0 liter of 5.0 m

olar salt solution?
Chemistry
1 answer:
AlladinOne [14]2 years ago
3 0
The concentration of the solution is 5.0 molar, which is 5.0 mole/L.  So in the 1.0 L of 5.0 molar KF salt solution, the moles of KF is 5.0molar*1.0L=5.0 mole. The molecular weight of KF is given in the question as 58.10 gram/mole, so the grams of KF is 58.10 gram/mole * 5.0 mole = 290.5 gram.
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Explanation : Given,

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Now we have to calculate the moles of MgO

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From the balanced reaction we conclude that

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So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

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\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

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Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

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V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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