6.3 mg of a boron hydride is contained in a flask of 385 mL at 25. degrees C and a pressure of 11 Torr.

A. determine the molar mass of the hydride.
B. which of the following hydrides is contained in the flask, BH3, B2H6, or B4H10

A. MM = 27.597 g/mol
B. B2H6

Here is my work, I might have gotten the answers the wrong way.

6.3 mg / 1000 = .0063g
385 mL / 1000 = .385 L
25 degrees C + 273 = 298 K
11 Torr / 760 = .0145 atm

.0145 atm x .385L = n (.08206 Latm/molK) 298K
n = 2.28x10^-4 moles

A. MM = .0063g / 2.28x10^-4mole = 27.597 g/mol
B. BH3 molar mass = 13.83 g
B2H6 = 27.67 g - this one is closest...
B4H10 = 53.32 g

A volume of 26.5 mL of nitrogen gas was collected in a tube at a temperature of 17° C and a pressure of 737 mm Hg. The next day the volume of the nitrogen was 27.1 mL with the barometer still reading 737 mm Hg. What was the temperature on the second day?

You did it right; I used 11/760 and left it in the calculator.

You have too many s.f. in that answer for A. Also, I don't get that exactly; the problem, I suspect, is that you rounded with intermediate numbers and I didn't.

B is ok.

A. Well, we can calculate the molar mass of the hydride by using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. But since we want the molar mass, we can rearrange the equation to solve for n, the number of moles.

n = (PV) / (RT)

Plugging in the values we have, P = 11 Torr, V = 385 mL (or 0.385 L), R is the gas constant which is 0.0821 L * atm / (mol * K), and T is 25 + 273.15 K = 298.15 K.

n = (11 Torr * 0.385 L) / (0.0821 L * atm / (mol * K) * 298.15 K)

Solving that equation should give us the number of moles, and then we can divide the mass by the number of moles to get the molar mass.

B. As for determining which hydride is in the flask, well, I can't really use my clown sense of humor for that. We'll have to look at the molar mass of each compound and compare it to the molar mass we calculated. The molar mass of BH3 is 13.83 g/mol, B2H6 is 27.67 g/mol, and B4H10 is 55.37 g/mol. Our calculated molar mass of 27.597 g/mol is closest to the molar mass of B2H6, so that's our hydride.

To determine the molar mass of the boron hydride, we can use the ideal gas law and the given information of mass, volume, temperature, and pressure.

A. To calculate the molar mass, we will first convert the mass of the boron hydride from milligrams (mg) to grams (g). This can be done by dividing the given mass of 6.3 mg by 1000:

Mass (g) = 6.3 mg / 1000 = 0.0063 g

Next, we need to calculate the number of moles of the boron hydride using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

First, we need to convert the given temperature of 25 degrees Celsius to Kelvin (K):

T (K) = 25 + 273.15 = 298.15 K

Next, we need to convert the given pressure of 11 Torr to atm by dividing it by 760 (since 1 atm = 760 Torr):

P (atm) = 11 Torr / 760 = 0.01447 atm

Now, we can rearrange the ideal gas law equation to solve for moles:

n = PV / RT

Substituting the values we calculated:

n = (0.01447 atm) * (0.385 L) / [(0.0821 L.atm/mol.K) * (298.15 K)]

n = 0.00195 mol

Finally, we can calculate the molar mass by dividing the mass of the boron hydride by the number of moles:

Molar Mass (g/mol) = 0.0063 g / 0.00195 mol = 27.597 g/mol

Therefore, the molar mass of the boron hydride is approximately 27.597 g/mol.

B. To determine which of the three given hydrides (BH3, B2H6, or B4H10) is contained in the flask, we need to compare the molar mass of the boron hydride we calculated (27.597 g/mol) with the molar masses of the possible hydrides.

The molar masses of the compounds are as follows:
- BH3 (boron trihydride) has a molar mass of approximately 13.83 g/mol.
- B2H6 (diborane) has a molar mass of approximately 27.67 g/mol.
- B4H10 (tetraborane) has a molar mass of approximately 53.58 g/mol.

Since the molar mass of the boron hydride we calculated is closest to the molar mass of B2H6 (27.67 g/mol), we can conclude that the hydride contained in the flask is B2H6 (diborane).