Question

Q.16. To prepare a solution with molality of 0.1 mol kg -1 of ethanol in 100g of water, the mass of ethanol needed is (molecular weight of ethanol = 46 g mol-1)​

Answer 1

Answer:

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Answer 2

Answer:

The mass of ethanol needed to prepare a solution with molality of 0.1 molkg⁻¹ is equal to 0.46g.

Explanation:

We have given,

The molality of ethanol solution = 0.1molkg⁻¹

The molecular mass of ethanol (CH₃CH₂OH) =46gmol⁻¹

The mass of water (solvent) = 100g = 0.1Kg

we know that molality of a solution is equal to the number of moles of solute divided by mass of solvent in kilograms.

Molality = moles of solute (ethanol)/mass of water(solvent)

Number of moles of ethanol = Molality ×mass of water

Moles of ethanol $= 0.1molkg^{-1}*\;0.1kg=0.01mol$

We know that moles = mass/molar mass

Mass of ethanol = Number of moles× molar mass of ethanol

Mass of ethanol = (0.01mol)×(46gmol⁻¹)=0.46g

Therefore, 0.46g mass of ethanol is required.