how many liters of gaseous hydrogen bromide at 97 degrees c and 0.953 atm will a chemist need if she wishes to prepare 3.50 L of 1.20 M hydrobromic acid?
How many mols does she need? That's M x L = ?
Plug that mols (n) into PV = nRT at the conditions listed and solve for V in L.
To solve this problem, we will use the Ideal Gas Law equation: PV = nRT, where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas (in Kelvin)
Given:
Volume of hydrobromic acid (HBr) required = 3.50 L
Concentration of hydrobromic acid (HBr) = 1.20 M
Temperature (T) = 97 degrees Celsius = 97 + 273.15 = 370.15 K
Pressure (P) = 0.953 atm
To find the number of moles (n) of HBr required, we can rearrange the equation as follows:
n = PV / RT
First, let's convert the given temperature to Kelvin by adding 273.15 to the Celsius temperature.
Now, substitute the given values into the equation:
n = (0.953 atm) * (3.50 L) / (0.0821 L·atm/(mol·K) * 370.15 K)
Simplifying the equation gives:
n = 0.0432 moles of HBr
Since one mole of HBr corresponds to one mole of gaseous hydrogen bromide (HBr), the number of moles of gaseous HBr required is also 0.0432 moles.
To convert moles of HBr to liters of gaseous HBr at the given conditions, we use the ideal gas law equation again, but this time solve for V:
V = nRT / P
Substituting the values:
V = (0.0432 moles) * (0.0821 L·atm/(mol·K)) * (370.15 K) / (0.953 atm)
Simplifying the equation gives:
V = 1.70 L
Therefore, the chemist will need approximately 1.70 liters of gaseous hydrogen bromide at 97 degrees Celsius and 0.953 atm to prepare 3.50 L of 1.20 M hydrobromic acid.