Question

3.0 molal sodium hydroxide (naoh) solution has a density of 1.110g/ml. the molarity of this solution is​

Answer 1

Answer:

• Molarity = 2.973 M

Steps:

Molality of the given NaOH soln: 3 molal

Density of the solution = 1.11 g/ml

According to Molality formula,

$\boxed{ Molality = \dfrac{ \text{No. of moles of solute}}{\text{Mass of Solvent in kg}}}$

Hence 3 molal NaOH solution would contain 3 moles of NaOH in 1 kg of water.

Hence Mass of NaOH = No of moles of NaOH × Molar Mass of NaOH

Mass of NaOH = 3 × ( 23 + 16 + 1 ) ⇒ 3 × 40 = 120g

Hence Mass of Solution = Mass of NaOH + Mass of Water

→ Mass of Solution = 120g + 1000g = 1120g

Now we know that,

⇒ Density of Soln. = Mass of Soln. / Volume of Soln.

Hence, Volume of Soln. is given as:

⇒ Volume of Soln. = Mass of Soln. / Density

⇒ Volume of Soln. = 1120g / 1.11 g/ml

⇒ Volume of Soln. = 1009 ml (approx.)

Hence Molarity of the solution can be given as:

$\implies Molarity = \dfrac{\text{No of moles of solute}}{\text{Volume of Soln. in liters}}$

Substituting in the formula we get:

$\implies \boxed{Molarity = \dfrac{3}{1.009} = 2.973\: M}$

Hence the molarity of the solution is 2.973 M.

Answer 2

Answer:

It is given that the density of the solution is 1.110 g/ml.

The concentration of sodium hydroxide solution is 3.0 molal.

This implies 1000 g of water has 3 moles of NaOH.

3 moles of NaOH = 3×40= 120g of NaOH

Mass of solution = 1000+ 120 = 1120g

volume of solution =

$\frac{mass}{density} = \frac{1120}{1.11} = 1009 \: ml$

Molarity =

$\frac{number \: of \: moles}{volume \: in \: l} = \frac{3}{ \frac{1009}{1000} } = 2.97 \: m$

Hence, the molarity of the solution is 2.97 M.