The molality of a 25.4 % (by mass) aqueous solution of phosphoric acid is .

**Further Explanation:**

The proportion of substance in the mixture is called concentration. The most commonly used **concentration terms** are as follows:

**1.** Molarity (M)

**2.** Molality (m)

**3.** Mole fraction (X)

**4.** Parts per million (ppm)

**5.** Mass percent ((w/w) %)

**6.** Volume percent ((v/v) %)

**Molality **is equal to the amount of solute that is dissolved in one kilogram of the solvent. It is denoted by m and its unit is moles per kilograms (mol/kg). The component present in **smaller quantity** is **solute **while that in **larger quantity** is known as a **solvent**. The solute gets itself dissolved in the solvent.

The formula to calculate the molality of solution is as follows:

…… (1)

Consider 100 g to be the mass of the sample. Phosphoric acid is 25.4 % by mass. So the mass of is 25.4 g.

The formula to calculate the mass of solution is as follows:

…… (2)

Rearrange equation (2) to calculate the mass of .

…… (3)

The mass of solution is 100 g.

The mass of is 25.4 g.

Substitute these values in equation (3).

Mass of is to be converted into kg. The conversion factor for this is,

So the mass of is calculated as follows:

The formula to calculate the moles of is as follows:

…… (4)

Substitute 25.4 g for the given mass and 97.99 g/mol for the molar mass of in equation (4).

Substitute 0.2592 mol for the amount of and 0.0746 kg for the mass of in equation (1).

Therefore the **molality of solution is 3.47 m.**

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**Answer details:**

**Grade:** Senior School

**Subject:** Chemistry

**Chapter:** Solutions

**Keywords:** concentration terms, molality, m, mol/kg, H3PO4, moles of H3PO4, mass of H2O, H2O, 25.4 %, 25.4 g, 100 g, 74.6 g, 0.2592 mol, 3.47 m.