Question

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Answer 1

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Answer 2

Solution :-

Given :-

Density of 2M NaOH = 1.28 g/cm³

Now as we know

$\small{Molarity(M) = \dfrac{No \: of \: moles \: of \: solute}{Volume\: of \: solution\: in \: liter}}$

$Molality(m) = \dfrac{No \: of \: moles \: of \: solute}{Mass\: of \: solvent\: in\: kg}$

Now let there is 1 liter of solvent.

Then

$\small{Molarity(M) = \dfrac{No \: of \: moles \: of \: solute}{Volume\: of \: solution\:in \: liter}}$

$\implies 2 = \dfrac{No \: of \: moles \: of \: solute}{1}$

$\implies No \: of \: moles \: of \: solute = 2\: moles$

Now as density = 1.28 g/cm³

Mass of solution

= 1.28 × 1litre

= 1.28 × 1000 cm³

= 1280 g

Also mass of solute

= 2 × molar mass

As mass of

Na = 23

H = 1

O = 16

So mass of solute

= 2 × (23 + 1 + 16)

= 2 × (40)

= 80 g

Now mass of solvent

= Mass of solution - Mass of solute

= 1280 - 80

= 1200 g

= 1.2 kg

So Molality (m)

$Molality(m) = \dfrac{No \: of \: moles \: of \: solute}{Mass\: of \: solvent\: in\: kg}$

$= \dfrac{2}{1.2}$

$= 1.666....$

$= 1.67 \: m$

So Molality (m)

$\Huge{\boxed{\sf{ = 1.67 \: m }}}$