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
Whoever Answers it correctly I will mark it brainliest​​

Answer 1

I really don't know but I want to help you yeah

Answer 2

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Yøur-Añswer♡}}}$

$\large\bf{\underline{\red{VERIFIED✔}}}$

$\bf\huge{\purple{╔════════♡︎═╗}}$

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,

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

$\huge{\textbf{{{\color{navy}{G}}{\red{i}}{\pink{ven♤࿐}}{\color{pink}{:}}}}}$

• 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

$\bf\huge{\pink{╚═♡︎════════╝}}$ .

$\huge\mathcal\colorbox{lavender}{{\color{b}{✿Thanks♡}}}$

$\boxed{I \:Hope\: it's \:Helpful}$