Question

Part A. You have 0.03720 mol of N2 gas at STP. What is its density? This is supposedly a trick question...please show your logic/reasoning so I can understand the problem. Thank you.

Part B.10.00 mL of sulfuric acid (H2SO4) is diluted to 100.00 mL. The diluted solution is titrated with 6.00 M KOH. The equivalence point is reached after the addition of 67.02 mL KOH. What is the concentration of the undilated solution?

Answer 1

Part A: no. of moles of N2 = 0.03720 mol = w/M

w = n*M = 0.03720*28 = 1.0416 gms

T = 298 K

P = 1 atm

From ideal gas equation PV = nRT

V = nRT/P = 0.03720mol*0.0821L.atm.K^-1*298/1 atm = 0.9101 L = 910 mL

density = mass/volume = 1.0416gms/910mL = 0.001144 g/mL

Part B: Inital volume of H2SO4 = 10 mL and concentration = M

if diluted 100mL, its molarity = 1/10

H2SO4 on titrating with6.00 M KOH consumed 67.02 mL

the reaction between H2SO4 and KOH is given by

H2SO4 + 2KOH -----> K2SO4 + 2H2O

1mole H2SO4 = 2mole KOH

The molarity of unknown acid can be found using the equation M1V1/n1 = M2V2/n2

M1 = molarity of H2SO4 = ?----------------------M2 = molarity of KOH = 6.00 M

V1 = volume of H2SO4 = 100mL-----------------V2 = volume of KOH consumed = 67.02 mL

n1 = mo. of moles of H2SO4 = 1------------------n2 = no. of moles of KOH = 2

M1 = M2V2*n1/V1*n2 = 6.00*67.02*1/100*2 = 2.01 M

The molarity of diluted H2SO4 = 2.01 M

The molarity of undiluted solution was M1V1 = M2V2

M1 =?, V1=10mL, M2 = 2.01 M, V2 = 100mL

M1 = M2V2/V1 = 2.01*100/10 = 20.1 M