Calculate the concentrations of all species present in a 0.25-M solution of ethylammonium chloride (C2H5NH3Cl).

Concentration of carbon = 0.074 M

Concentration of hydrogen = 0.025 M

Concentration of nitrogen = 0.043 M

Concentration of chlorine = 0.108 M

Explanation:

Molar mass of C2H5NH3Cl = 81.5g/mole

Concentration of C2H5NH3Cl = 0.25 M

Concentration of C = 24/81.5*0.25 = 0.074 N

Concentration of H = 8/81.5*0.25 = 0.025 M

Concentration of N = 14/81.5*0.25 = 0.043 M

Concentration of Cl = 35.5/81.5*0.25 = 0.108 M

The molar mass of C2H5NH3Cl = (2*12) + (5*1) + (14) + (3*1) + 35.5 = 24+5+14+3+35.5 = 81.5

for carbon:

24/81.5 x 0.25M = 0.074M

for H:

8/81.5 x 0.25M = 0.0245M

for N:

14/81.5 x 0.25M = 0.0429M

for Cl:

35.5/81.5 x 0.25M = 0.1088M