Calculate the molality of a 25.4% (by mass) aqueous solution of phosphoric acid (H3PO4).

W=0.254 (25.4%)

M (H₃PO₄) = 98.0 g/mol

m - the mass of the solution

m (H₃PO₄) = mw

n (H₃PO₄) = m (H₃PO₄) / M (H₃PO₄)

m (H₂O) = m (1-w)

the molality is:

Cm=n (H₃PO₄) / m (H₂O)

Cm=mw/[M (H₃PO₄) m (1-w) ]=w/[M (H₃PO₄) (1-w) ]

Cm=0.254/[98 * (1-0.254) ]=0.003474 mol/g=3.474*10⁻³ mol/g=3.474 mol/kg