Question

30 g. of acetic acid is dissolved in 1 dm of a solvent. The molality of the solution will be (Given density of solvent = 1.25 g. cm) a) 0.4 b) 0.3 c) 0.5 d) 0.45​

Answer 1

Correct question :-

30 g of acetic acid is dissolved in 1 dm³ of a solvent. The molality of the solution will be ? (Given density of solvent = 1.25 g/cm³)

Answer :-

Molality of the solution is 0.4 m [Option.a]

Explanation :-

We have :-

→ Mass of acetic acid = 30 g

→ Volume of solvent = 1 dm³ = 1000 cm³

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Molar mass of acetic acid is 60 g/mol . So, number of moles of solute will be :-

= Given Mass/Molar mass

= 30/60

= 0.5 mole

Mass of solvent :-

= Volume × Density

= 1000 × 1.25

= 1250 g

= 1.25 kg

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Molaity of the solution :-

= Moles of solute/Mass of solvent in kg

= 0.5/1.25

= 50/125

= 0.4 m

Answer 2

• Mass of solute=30g

Molar mass of acetic acid

$\\ \sf\longmapsto CH_3COOH$

$\\ \sf\longmapsto 12u+4(1u)+2(16u)+12u$

$\\ \sf\longmapsto 60u$

$\\ \sf\longmapsto 60g/mol$

No of moles:-

$\\ \sf\longmapsto No\:of\;moles=\dfrac{Given\:mass}{Molar\:mass}$

$\\ \sf\longmapsto No\:of\:moles=\dfrac{30g}{60g/mol}$

$\\ \sf\longmapsto No\:of\;moles=0.5mol$

• volume of solvent=1dm^3=1000cm^3
• Density=1.25g/cm^3

Mass of solvent:-

$\boxed{\sf Density=\dfrac{Mass}{Volume}}$

$\\ \sf\longmapsto Mass=Density\times Volume$

$\\ \sf\longmapsto Mass=1000\times 1.25$

$\\ \sf\longmapsto Mass=1250g$

$\\ \sf\longmapsto Mass=1.25kg$

Molality:-

$\boxed{\sf Molality=\dfrac{Moles\:of\:solute}{Mass\:of\:solvent\:in\:gram}}$

$\\ \sf\longmapsto Molality=\dfrac{0.5}{1.25}$

$\\ \sf\longmapsto Molality=0.4m$