Question

Calculate the density of a sample of 1.00 mole of nh3 at 793mmhg and -9.00 c

Answer 1

0.164 g/L is the density of a sample of 1.00 mole of $NH_3$ at 793mmhg and -9.00 degrees celcius.

What is density?

Density is the mass of a unit volume of a material substance. The formula for density is d = $\frac{M}{V}$, where d is density, M is mass, and V is volume.

Given data:

n = 1.00 mole

P=793 mm hg =1.04342 atm

T=-9.00 degree celcius = -9.00 + 273= 264 K

V=?

Using Ideal Gas Law equation:  

PV = n R T      

R = gas constant = 0.082057 L-atm/(mol-K)

(1.04342 atm)(V) = 5 X 0.082057 L-atm/(mol-K)  X 264 K

V = 103.67 Liters

Now calculate density:

Mole weight of $NH_3$ = 1.00 mole

So, the mass of $NH_3$ = 17.031 g

Density =  $\frac{mass}{volume}$  

Density =  $\frac{17.031 g}{ 103.67 Liters}$  

= 0.164 g/L

Hence, 0.164 g/L is the density of a sample of 1.00 mole of $NH_3$ at 793mmhg and -9.00 degrees celcius.

Learn more about the density here:

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