Question

Dissolving 120gr of urea in 1000gr of water gave a solution of density 1.12g/ml. Find the molarity of the solution?

Answer 1

$\bigstar$ Answer:

$\checkmark$ Given:

⇒ Weight of urea (CH₄N₂O) = 120 g

⇒ Weight of water (H₂O) = 1000 g

⇒ Density (d) = 1.12 g/ml

$\checkmark$ To Find:

⇒ Molarity (M) of the solution

$\checkmark$ Solution:

$\dagger$ We know that,

$\star$ Density (d)

$\green{\boxed{\sf Density(d)=\dfrac{Mass(M)}{Volume(V)} }}$

Here,

We need to find the Volume (V)

Therefore,

$\red{\implies \sf Volume(V)= \dfrac{Mass(M)}{Density(D)}}$

Here,

→ In this Solution,

The Mass of Solution is 1120 grams

Since,

⇒ Mass of Urea = 120 g

⇒ Mass of Water = 1000 g

So,

Mass of the Total Solution is

∴ Mass (M) = (120 + 1000) g

Hence,

⇒ The Mass of Solution (M) is 1120 grams

Given that,

⇒ Density (d) = 1.12 g/ml

Therefore,

Substituting the above values in the Formula

We get,

$\implies \sf Volume(V)= \dfrac{1120}{1.12}\;\;ml$

$\blue{\implies \sf Volume(V)= 1000\;\;ml}$

$\implies \sf Volume(V)= 1\;\;L$

∵ 1 L = 1000 ml

$\dagger$ We know that,

$\star$ Molarity (M)

$\pink{\boxed{\sf Molarity (M)=\dfrac{W}{GMW} \times \dfrac{1000}{V(ml)}}}$

Given that,

W = Weight of urea dissolved = 120 g

⇒ W = 120 g

GMW = Molecular Weight of urea (CH₄N₂O) = (12 + 4 + 28 +16)g

⇒ GMW = 60 g

And,

We Found that,

⇒ Volume (V) = 1000 ml

Therefore,

Substituting the above values in the Formula

We get,

$\implies \sf Molarity (M)=\dfrac{120}{60} \times \dfrac{1000}{1000}$

$\implies \sf Molarity (M)=\dfrac{120}{60} \times 1$

$\implies \sf Molarity (M)=\dfrac{2}{1}$

$\purple{\implies \sf Molarity (M)=2 \ M}$

Hence,

The Molarity of the Solution is 2

⇒ Molarity (M) = 2 M