Question

How many grams of alcl3 are required to make a 2.25m solution in 30.0 g of water?

Answer 1

We know that, Molality = $\frac{\text{weight of solute (g)}}{\text{molecular weight X Weight of solvent (Kg)}}$

Given: Molality of solution = 2.25 m
Weight of solvent = 30 g = 0.030 kg
Molecular weight of AlCl3 = 133.34 g/mol

Therefore we have, 2.25 = $\frac{\text{weight of solute (g)}}{\text{133.34 X 0.030}}$
∴, weight of solute (g) = 9.00 g

Thus, 9.00 g of AlCl3 is required to make a 2.25m solution in 30.0 g of water

Answer 2

Molality is the ratio of the moles to the mass of the solvent. In a 2.25 m solution,  9.00 g of AlCl₃ is required.

What is molality?

Molality is the number of moles of solute dissolved in a kilogram of the solvent to make the solution.

Given,

Molality (m) = 2.25 m

Weight of water (solvent) = 30.0 gm

The molecular weight of AlCl₃ = 133.34 g/mol

The weight of solute is calculated as:

m = mass of solute ÷ (molar mass of solute × mass of solvent (kg))

2.25 = weight of solute (g) ÷ (133.34 × 0.030)

= 9.00 gms

Therefore, 9.0 gms of solute are present in 30 gm of solvent.

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