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larisa [96]
1 year ago
5

What is the molality of a solution prepared by dissolving 3. 41 g of c6h12o6 in 85 ml of water:__________

Chemistry
1 answer:
vovikov84 [41]1 year ago
7 0

When 3. 41 g of C6H12O6 is dissolved in 85 ml of water the molality is 2.2m.

<h3>What is molality? </h3>

Molality is the amount of a substance dissolved in a certain mass of solvent.

Molality = moles of solute/ mass of solvent (kg)

Given,

Mass of solute = 3.14g

Molar mass of solute = 180 g

Mole of solute= given mass/molar mass

= 3.14/180

= 0.0189 mol.

Volume of solvent = 85 ml

Density of water = 1 g/cm3

Density = mass/ volume

mass = density × volume

= 1× 85

= 85g

Molality = (0.0189/ 85) × 1000

= 2.2m

Thus we find that the molality of given solution is 2.2m.

learn more about molality:

brainly.com/question/26921570

#SPJ4

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The equilibrium reaction is

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Given:

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Answer:

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For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:

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The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):

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