Question

calculate the molality of a solution when 1.5 moles is added to 675 ml of solvent (density of solvent = 1g/mol).
please give step by step explanation​

Answer 1

Answer:

Only if we consider the solvent in this question is water ,we can consider the mass of solvent to be 675mL =675 g

since for water:

1L =1 kg

And thus  

Molality of solution = Moles of solute / kg of solvent

                           =1.5/0.675 = 2.2 m

Explanation:

Answer 2

Answer:

The molality of the solution is 2.22m.

Explanation:

Given data is,

No.of moles of solution=1.5mol

Volume of the solvent(V)=675ml

Density of solvent(ρ)=1g/cc

The density of the solvent is the ratio of its mass to volume.Therefore,the mass of the solvent is the product of its density and volume.

Hence we write

$m=V\rho=675ml\times1g/cc=675g\\=0.675kg$

We know that,

The molality of a solution is the ratio of number of moles to weight of solvent(in kilograms). Therefore,

$m=\frac{n}{w_{solvent}(in\:kg)}$

Substituting the known values in the above relation,we get

$m=\frac{1.5}{0.675}=2.22m$

Therefore,

The molality of the solution is 2.22m.

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