M(NH₄Cl) = 14.00 + (4 x 1.008) + 35.45 = 53.48 g mol^-1

n(NH₄Cl) = 3.50g / 53.48 g mol^-1 = 0.06545 mol
c(NH₄Cl) = 0.06545 mol / 0.250 L = 0.262 mol L^-1