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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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Explanation :

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$\log \frac{I_o}{I}=(855dm^3mol^{-1}cm^{-1})\times (3.25\times 10^{-3}mol.dm^{-3})\times (0.25cm)$

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