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
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Answer 1

Answer:

The molar mass of NaOH is 40 g/mol.

The number of moles of NaOH is equal to the ratio of its mass to molar mass.

The number of moles of NaOH =

40g/mol

40g

=1mol

The molarity of NaOH solution is the ratio of number of moles of NaOH to total volume of solution in L.

M=

1L

1mol

=1M

Answer 2

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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

Volume = 200 mlDensity = 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

Total mass of solution = 250 gMass 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,

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

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Moles of Solute = 1 mole

Moles of Solute = 1 moleMass 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

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