Calculate the molality of a 27.0% (by mass) aqueous solution of nitric acid.

Answer is: molality is 5.87 m.

ω(HNO₃) = 27.0% = 0.27.

If we use mass of solution 100 grams:

m(solution) = 100 g.

m(HNO₃) = ω(HNO₃) · m(solution).

m(HNO₃) = 0.27 · 100 g.

m(HNO₃) = 27 g.

n(HNO₃) = m(HNO₃) ÷ M(HNO₃).

n(HNO₃) = 27 g ÷ 63 g/mol.

n(HNO₃) = 0.428 mol.

m(H₂O) = 100 g - 27 g.

m(H₂O) = 73 g ÷ 1000 g/kg.

m(H₂O) = 0.073 kg.

b(solution) = n(HNO₃) ÷ m(H₂O).

b(solution) = 0.428 mol ÷ 0.073 kg.

b(solution) = 5.87 mol/kg = 5.87 m.

the answer is   c.) 1575 j.

explanation:

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.