Question

when 1.570 grams of the compound is vaporizedar 300 degrees celius and 1 atmosphere, the gas occupies a volume of 577 milliliters. what is the molar mass of the compound based on the result?

Answer 1

Answer:

128 g/mol

Explanation:

The molar mass is how much a mol of the substance weight.

In this question the weight is given (1.570 g).

To find the number of moles in the sample, we use the ideal gas law: PV=RnT

P= 1atm

V= 577mL= 0.577L

R (universal gas constant)= 0.08206($\frac{atm*L}{mol*K}$)

n= number of moles

T= 300°C= 573.15K

Solving for n:

n= $\frac{1 atm*0.577L}{0.08206\frac{atm*L}{mol*K}*573.15}$ k

n=0.0123mol

Molar mass=$\frac{1.570g}{0.0123 mol}$

Molar mass=128$\frac{g}{mol}$

Answer 2

Answer:

128 g/mole.

Explanation:

To solve this question we will need the formula for the ideal gas law and the formula for Calculating the numbers of mole stoichiometrically and they are both given below respectively.

PV = nRT ----------------(ideal gas law).

where P = pressure, V= volume, n= number of moles, R= gas constant, T= temperature.

n = m / Mm. ------------------- *** ( formula for Calculating the number of moles).

Where n= number of moles,m= mass and Mm= molar mass.

We will the. Substitute the equation *** into the ideal gas law equation to give;

PV = (m/Mm) × R × T.

Since, we are looking for the molar mass, then the equation above will be;

Mm = ( m × R × T ) ÷ P × V.

==> Mm = (1.570 × 0.0821 × 573) ÷ 1 × 577 × 10^-3.

==> Mm = 73.857981÷ 0.577.

Mm= 128 grams per mole.