Question

7. Calculate the pressure that CCl₄ will exert at 40 °C if 1.00 mole occupies 28.0 L assuming that CCl₄ behaves like a real gas. (Values for the van der Waals constants are a = 19.7 L²atm/ mol², b = 0.1281 L/mol.)

8. HCN can be formed by the reaction below:

NaCN(s) + HCl(aq) à HCN(g) + NaCl(aq)

What mass of NaCN is required to make 2.25 L of HCN at 30°C and 748 mmHg? (3 points)

Answer 1

7.

n = 1 mole

T   = 40 + 273   = 313K

V = 28L

a = 19.7 L²atm/ mol², b = 0.1281 L/mol.

(P + an^2/v^2)(v-nb) = nRT

(P + 19.7(1/28)^2)(28-1*0.1281) = 1*0.0821*313

( P + 0.025)(27.8719)   = 25.6973

P   = 0.9atm >>>>answer

8.

V = 2.25L

T = 30 + 273 = 303K

P = 748/760   = 0.9842atm

PV   = nRT

n   = PV/RT

      = 0.9842*2.25/(0.0821*303)

       = 0.09moles

NaCN(s) + HCl(aq) -------------> HCN(g) + NaCl(aq)

1 mole of HCN produced from 1 mole of NaCN

0.09mole of HCN produced from 0.09 moles of NaCN

mass of NaCN   = no of moles * gram molar mass

                        = 0.09*49

                           = 4.41g