Question

How many millilitres of water are required to dissolve 25 g of urea in order to produce a solution that is 1.65 m?

“m” mentioned above is molality.

Answer 1

Solution :-

We have got :-

Mass of urea = 25 g

Required Molality = 1.65m

Now as we are known that :-

Density of water = 1000 kg/m³

Formula for urea = CH₄N₂O

Now Mass of :-

Carbon atom = 12 u

Oxygen atom = 16 u

Nitrogen atom = 14 u

Hydrogen atom = 1 u

Now :-

Molecular Mass of Urea

= (1)12 + (4)1 + (2)14 + 16

= 12 + 4 + 28 + 16

= 60 u

Then molar mass = 60 g

Now as we know :-

$\sf{No. \: of \: Moles = \dfrac{Given\:Mass}{Molar\:Mass}}$

$\sf{No. \: of \: Moles = \dfrac{25}{60}}$

$\sf{No. \: of \: Moles = \dfrac{5}{12}}$

Now we know that :-

$\sf{Molality = \dfrac{Moles\: of \:Solute}{Mass\: of \: Solvent\:(kg)}}$

Let the weight of water be x kg

So

$\rightarrow 1.65 = \dfrac{\dfrac{5}{12}}{x}$

$\rightarrow 1.65x = \dfrac{5}{12}$

$\rightarrow x = \dfrac{\dfrac{5}{12}}{1.65}$

$\rightarrow x = \dfrac{5}{19.8}$

$\rightarrow x = 0.2525...$

$\rightarrow x = 0.252$

So Mass of water = 0.252 kg

So volume of water

= 0.252 ÷ 1000 kg/m³

= 0.000252 m³

= 0.252 liters

Or

= 252 ml

So we require 252 ml of water

Answer 2

Answer:

252 ml or 0.252 l of water is required

Explanation:

We have got :-

Mass of urea = 25 g

Required Molality = 1.65m

Now as we are known that :-

Density of water = 1000 kg/m³

Formula for urea = CH₄N₂O

Now Mass of :-

Carbon atom = 12 u

Oxygen atom = 16 u

Nitrogen atom = 14 u

Hydrogen atom = 1 u

Now :-

Molecular Mass of Urea

= (1)12 + (4)1 + (2)14 + 16

= 12 + 4 + 28 + 16

= 60 u

Then molar mass = 60 g

Now as we know :-

Now we know that :-

Let the weight of water be x kg

So

So Mass of water = 0.252 kg

So volume of water

= 0.252 ÷ 1000 kg/m³

= 0.000252 m³

= 0.252 liters

Or

= 252 ml

So we require 252 ml of water