Question

Calculate the density of nitrogen at 695 Torr and –15°C. A. 1.08 g/L B. 9.07 g/L C. 2.42 g/L D. 1.21 g/L E. 0.605 g/L Show steps please.

Answer 1

Given, P = 695 torr = 695 torr x (1 atm/760 torr) = 0.914 atm

T = -150C = (-15+273) K = 258 K

From ideal gas law we have,

PV = nRT

Again, n = m/M , where m = mass and M = molar mass

Also, m(mass0 = d(density) x v(volume)

So the aboe realtion becomes as:

PV = (d*V/M)*RT

Rearrange the formula

d = P*V*M/V*RT

=> d = PM/RT

Now, molar mass(M) for nitrogen(N2) gas is = 2 x 14 = 28 g/mol

and R (gas constant) has the value 0.08206 L atm/mol.K

so, d = 0.914 atm x 28 gmol-1 / 0.08206 Latm mol-1K-1 x 258 K

=> d = 25.592/21.17148 g/L

=> d = 1.208 g/L =1.21 g/L

Hence, the answer is D) 1.21 g/L