Question

A certain amount of NaOH is dissolved in certain kilograms of solvent and molality of the solution is '0.5 m' . When the same amount of NaOH is dissolved in "100 grams of less solvent than initial" then molality becomes '0.625 m'.

Determine the amount of NaOH and the initial mass of solvent .

Proper explanation expected. Best of luck guys ! :)​​

Answer 1

Answer:

10grams

we all know that molality =moles(solute)/weight (solvent in kg)

Let moles of NaOH be x and weight of Solvent be Ykg

further solution in attachment

ALSO 2X=Y

X=.25

than y=.5kg

Y=500grams

Answer 2

⇝ Given :-

• A certain amount of NaOH is dissolved in certain kilograms of solvent and molality of the solution is '0.5 m.

• When the same amount of NaOH is dissolved in "100 grams of less solvent than initial" then molality becomes '0.625 m.

⇝ To Find :-

• The amount of NaOH

• The initial mass of solvent .

⇝ Solution :-

Let,

• Number of moles of NaOH dissloved = x moles

• Initial Mass of Solvent = y kg.

❒ We know molalaity (m) of solution is :

$\text m = \dfrac{ \text{No. of moles of Solute}}{ \text{Mass of Solvent in kg}}$

★ In 1st Condition Given That ;

When certain amount of NaOH is dissolved in certain kilograms of solvent and molality of the solution is '0.5 m.

❒ Hence, Using Formula of Molality :

$0.5 = \dfrac{ \text x}{ \text y}$

$: \longmapsto0.5 \text{y} = \text x$

✏ Multiplying Both Sides by 2

$\large: \longmapsto \bf y = 2x \: \: - - - (1)$

★ In 2nd Condition Given That ;

When the same amount of NaOH is dissolved in "100 grams of less solvent than initial" then molality becomes '0.625 m.

❒ Hence, Using Formula of Molality :

$\dfrac{ \text x}{\text y - 100\text g} = 0.625$

$: \longmapsto \dfrac{\text x}{\text y - \frac{100}{1000}\text {kg} } = 0.625$

$:\longmapsto \dfrac{\text x}{\text y - 0.1} = 0.625$

$: \longmapsto \text x = 0.625\text y - 0.0625$

$: \longmapsto0.625\bf y - x = 0.0625 \: \: - - - (2)$

✏ Putting (1) in (2) :

$: \longmapsto0.625(2 \text x) -\text x = 0.0625$

$:\longmapsto 1.25\text x - \text x = 0.0625$

$:\longmapsto0.25\text x = 0.0625$

$:\longmapsto\text x = \cancel\dfrac{0.0625}{0.25}$

$\Large \purple{ : \longmapsto \underline{\boxed{ \bf x = 0.25}}}$

✏ Putting Value of x in (1) :

$: \longmapsto \text y = 2 \times 0.25$

$\Large \purple{ : \longmapsto \underline{\boxed{ \bf y = 0.5}}}$

As

• No. of Moles of NaOH = x = 0.25 moles ;

We Know,

• Molar Mass of NaOH = 40 g

So,

$\text{Amt. of NaOH = x} \times \text{molar mass}$

$= 0.25 \times 40$

Hence,

$\underline{\pink{ \underline{\bf Amt. \: of \: NaOH =10 \: g}}}$

And,

Intial Mass of Solvent = y kg = 0.5 kg.

Hence,

$\underline{\pink{ \underline{\bf Initial \: Mass \: of \: Solvent =500 g}}}$