Question

4g sodium hydroxide is dissolved in 1 decilitre of solution. Find its molality if the density of the solution is 1.038g/cc?

Answer 1

Answer:

$\large Molality=1.002 \ mol/g$

Explanation:

Molality :

Molality is define as numbers of moles of solute present in unit

mass of solvent in kg.

Given mass = 4 g

Molar mass of NaOH = 40 u

Volume = 1 decilitre =  100 mL

Density = 1.038 g / cc

First let us find mass of solution

we know that

$\large density = \dfrac{mass}{volume} \\\\\\\large mass = 1.038\times100 \ g\\\\\\\large mass = 103.8 \ g$

Now mass of solvent = mass of solution - mass of solute

Mass of solvent = 103.8 - 4 = 99.8 g

Now number of moles = given mass / molar mass

Number of moles = 4 / 40 = 0.1 mole

We know formula for molality

$\large Molality=\dfrac{Numbers \ of \ moles \ of \ solute}{Mass \ of \ solvent \ (g)}\times1000$

Put the values here

$\large Molality=\dfrac{0.1}{99.8}\times1000\\\\\\\large Molality=\dfrac{100}{99.8}\\\\\\\large Molality=1.002 \ mol/g$

Thus we get answer.

Answer 2

Given mass (m) = 4 g

Molar mass of NaOH (m.m) = 40 g

Number of moles (n) of NaOH = m/m.m

=> 4/40 = 0.1 mol

Density = 1.038 g/cc

Volume = 100 ml

____________ [GIVEN]

• We have to find it's molality (m).

______________________________

Now..

Molality = $\dfrac{Number\:of\:moles\:of\:solute}{Mass \:of\:solvent}$

• Number of moles of solute = $\frac{Given\:mass}{Molar\:mass}$

• Mass of solvent = $\frac{Mass\:of\:solvent}{1000}$

_____________________________

Density = $\dfrac{Mass}{Volume}$

1.038 = $\frac{Mass}{100}$

Mass = 1.038 × 100

=> 103.800 g

______________________________

Mass of solvent = Mass of NaOH (solution) - Mass of solute

=> 103.800 - 4 = 99.8 g

______________________________

Molality (m) = $\dfrac{0.1}{99.8}$ × 1000

=> 100/99.8

=> 1.002 molal or mol/g

______________________________

1.002 molal or mol/g is the Molality.

___________ [ANSWER]

______________________________