Question

how many moles of gaseous boron trifluoride, bf3, are contained in a 4.3410-l bulb at 788.0 k if the pressure is 1.220 atm? how many grams of bf3?

Answer 1

Given situation is ,

( P) Pressure of BF3 gas is 1.220 atm

( V) Volume of BF3 is 4.3410L

( T) Temperature of BF3 is 788.0 K

Now, assuming this as an ideal gas,

Let us apply ideal gas equation,

PV = nRT

n = PV/RT = 1.220 atm * 4.3410L / 0.0821 L atm k-1 mol-1 * 788 K

n = 0.08186 moles of BF3 are present.

Now, we know molar mass of BF3 = 67.82 gmol-1

i.e. total mass of BF3 present = Number of moles * molar mass

                                                = 0.08186 * 67.82g mol-1

                                                = 5.552 g of BF3 is present.

What do you mean by ideal gas equation?

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as

PV = nRT

Where,

P is the pressure of the ideal gas.

V is the volume of the ideal gas.

n is the amount of ideal gas measured in terms of moles.

R is the universal gas constant.

T is the temperature.

To know more about ideal gas equation from the given link:

https://brainly.com/question/20348074

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