If you want to prepare 5 liters of a 0.35m solution of nh4cl, how many grams of salt

Answer is: mass fo ammonium chloride is 93.625 grams.

V (NH₄Cl) = 5 L.

c (NH₄Cl) = 0.35 M.

n (NH₄Cl) = V (NH₄Cl) · c (NH₄Cl).

n (NH₄Cl) = 5 L · 0.35 mol/L.

n (NH₄Cl) = 1.75 mol.

M (NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.

m (NH₄Cl) = n (NH₄Cl) · M (NH₄Cl).

m (NH₄Cl) = 1.75 mol · 53.5 g/mol.

m (NH₄Cl) = 93.625 g.