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3 years ago

What is the molality of a solution prepared by dissolving 150.0 g C6H12O6 in 600.0 g of H2O?

Chemistry

2 answers:

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

50/36 = 25/18

Explanation:

Solution at attachment box

Molality = mole of dissolvable (this question glucose) / kg of water

Masja [62]3 years ago

3 0

Answer:

Molality = 1.38 mol/Kg

Explanation:

Given data:

Molality of solution = ?

Mass of C₆H₁₂O₆ = 150.0 g

Mass of water = 600.0 g (600 g ×1 kg/1000 g= 0.6 Kg)

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 150.0 g/180.16 g/mol

Number of moles = 0.83 mol

Molality:

Molality = moles of solute / kg of solvent

Molality = 0.83 mol /0.6 Kg

Molality = 1.38 mol/Kg

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Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

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<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

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$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

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Learn more about oxidation numbers here:-

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