Question

$A \: solution \: is \: prepared \: by \\ dissolving \: 2 \: g \: of \: NaOH \\ in \: 8 \: g \: of \: water. \: Calculate \\ the \: mass \: percent \: and \: molality \\ of \: the \: solution.$

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Answer 1

HEY MATE!!!

PLZZ REFER THE ATTACHMENT

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Answer 2

Solution :-

Now we have got the mass of

Water = 8 g

NaOH = 2 g

Now no. of Moles of NaOH

= Given mass ÷ molar mass

= 2 ÷ 40

= 0.05 moles

As we know that the mass %

$= \dfrac{\textsf{mass of substance}}{\textsf{Total mass of solution}} \times 100$

As we know Molality

$= \dfrac{\textsf{Moles of Solute}}{\textsf{Mass of Solvent in kg}}$

Now

Mass % of NaOH

$= \dfrac{2}{10} \times 100$

$= 20 \%$

Mass % of Water

$= \dfrac{8}{10} \times 100$

$= 80\%$

Molality of NaOH

$= \dfrac{ 0.05}{8 \times 10^{-3}\: kg}$

$= \dfrac{0.05 \times 10^3}{8}$

$= \dfrac{50}{8}$

$= 6.25 \: m$