Question

Calcular la m y M de una concentración Porcentual 45% m/m, la cual tiene una D solución 1.25 g/ml y una masa de sto de 125g.HN03

Answer 1

Explanation:

Calculate the "m" and "M" of a percentage concentration of 45% m / m, which has a solution D 1.25 g / ml and a mass of 125g.

Given data is:

The mass % of the solution is 45%.

The density of the solution is 1.25 g/mL.

The mass of the solution is 125 g.

Molarity of solution is:

$M=\frac{number of moles of solute}{volume of solution}$

The volume of the solution is:

$Volume=\frac{mass}{density} \\=125 g / 1.25 g/ml\\=100 mL$

From mass% of the solution, the mass of solute HNO_3 can be calculated.

$mass percent=\frac{mass of solute}{ mass of solution} \\45=(mass of HNO_3/125 g)x100\\\\mass of solute = 56.25g$

The number of moles of HNO3 is :

$n_H_N_O__3=\frac{m_H_N_O__3}{M_H_N_O__3}$

$=56.25 g/63.00g/mol\\\\=0.893mol$

The molarity is:

$M=n/V\\M=0.893 mol/0.1L\\M=8.93 M$

The molality is:

$m=n/mass of solvent in kg\\ =0.893 mol /(0.125-0.05625)kg\\ = 12.9 m$