Question

40 gram of NaOH is present in 200 ml of aqueous solution of of density 1.25 gram per ml . Find molality of the solution . Given NaOH is the solute​

Answer 1

Answer :-

Molality of the solution is 4.76 m .

Explanation :-

We have :-

→ Mass of NaOH = 40 gram

→ Volume of the solution = 200 mL

→ Density of the solution = 1.25 g/mL

→ Molar mass of NaOH = 40 g/mol

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Firstly, let's calculate the number of moles of NaOH (solute) .

= Given Mass/Molar mass

= 40/40

= 1 mole

Mass of the solution :-

= Volume × density

= 200 × 1.25

= 250 g

Mass of solvent :-

= Mass of solution - Mass of solute

= (250 - 40)g

= 210 g

= 0.21 kg

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Molality of a solution :-

= Moles of solute/Mass of solvent in kg

= 1/0.21

= 100/21

= 4.76 m

Answer 2

Answer:

Given :-

• 40 gram of NaOH is present in 200 ml of aqueous solution of density 1.25 gram/ml.
• NaOH is the solute.

To Find :-

• What is the molality of the solution.

Solution :-

First, we have to find the numbers of moles in solute NaOH :

As we know that,

$\mapsto \sf\boxed{\bold{\pink{Number\: of\: Moles =\: \dfrac{Mass}{Molar\: Mass}}}}$

Given :

• Mass = 40 gram
• Molar mass = 40.0 g/mol

According to the question by using the formula we get,

$\implies \sf Number\: of\: moles_{(NaOH)} =\: \dfrac{40}{40.0}$

$\implies \sf\bold{\purple{Number\: of\: moles_{(NaOH)} =\: 1\: mole}}$

Now, we have to find the mass of NaOH :

As we know that,

$\mapsto \sf\boxed{\bold{\pink{ \rho =\: \dfrac{m}{v}}}}$

where,

• $\sf \rho$ = Density
• m = Mass
• v = Volume

Given :

• Volume = 200 ml
• Density = 1.25 gram/ml

According to the question by using the formula we get,

$\implies \sf 1.25 =\: \dfrac{m}{200}$

By doing cross multiplication we get,

$\implies \sf m =\: 200 \times 1.25$

$\implies \sf m =\: 200 \times \dfrac{125}{100}$

$\implies \sf m =\: \dfrac{250\cancel{00}}{1\cancel{00}}$

$\implies \sf m =\: \dfrac{250}{1}$

$\implies \sf\bold{\purple{m =\: 250\: gram}}$

Now, we have to find the mass of solvent :

As we know that,

$\footnotesize\mapsto \sf\boxed{\bold{\pink{Total\: mass\: of\: solution =\: Mass\: of\: Solvent + Mass\: of\: Solute}}}$

Given :

• Total mass of solution = 250 g
• Mass of solute = 40 g

According to the question by using the formula we get,

$\implies \sf 250 =\: Mass\: of\: Solvent + 40$

$\implies \sf 250 - 40 =\: Mass\: of\: Solvent$

$\implies \sf 210 =\: Mass\: of\: Solvent$

$\implies \sf\bold{\purple{Mass\: of\: Solvent =\: 210\: g}}$

Now, we have to convert mass of solvent g to kg :

$\implies \sf Mass\: of\: Solvent =\: 210\: g$

$\implies \sf Mass\: of\: Solvent =\: \dfrac{21\cancel{0}}{100\cancel{0}}\: kg\: \: \bigg\lgroup \sf\bold{1\: g =\: \dfrac{1}{1000}\: kg}\bigg\rgroup$

$\implies \sf Mass\: of\: Solvent =\: \dfrac{21}{100}\: kg$

$\implies \sf\bold{\green{Mass\: of\: Solvent =\: 0.21\: kg}}$

Now, we have to find molality of a solution :

As we know that,

$\mapsto \sf\boxed{\bold{\pink{Molality =\: \dfrac{Mass\: of\: Solute}{Mass\: of\: Solvent (kg)}}}}$

Given :

• Moles of Solute = 1 mole
• Mass of Solvent (kg) = 0.21 kg

According to the question by using the formula we get,

$\longrightarrow \sf Molality =\: \dfrac{1}{0.21}$

$\longrightarrow \sf Molality =\: \dfrac{1}{\dfrac{21}{100}}$

$\longrightarrow \sf Molality =\: \dfrac{1}{1} \times \dfrac{100}{21}$

$\longrightarrow \sf Molality =\: \dfrac{100}{21}$

$\longrightarrow \sf\bold{\red{Molality =\: 4.76\: m}}$

$\therefore$ The molality of a solution is 4.76 m .