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svetoff [14.1K]
3 years ago
11

a solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. what is the freezing point depression of

the solvent if the freezing point constant -1.86 C/m ?
Chemistry
1 answer:
Lesechka [4]3 years ago
3 0
The freezing point depression of the solution or pure substance that is added with the solvent is calculated through the equation,
    
  ΔTf = Kfm

where ΔT is the freezing point depression, Kf is the constant for water given to be -1.86°C/m and m is the molality of the solution. 

Molality is calculated through the equation,
    
    m = number of moles solute/ kg of solvent

Calculation of molality is shown below.
    
   m = (21.5 g C6H12O6)(1 mol/180 g) / (0.255 kg)
     m = 0.468 molal

The freezing point depression is then,
    
   ΔTf = (-1.86°C/m)(0.468 m) = -0.87°C

<em>Answer: -0.87°C</em>
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3 years ago
I need help with homework
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The term chemical reaction has to do with the combination of substances to give products. We know that reactants and products must follow the law of conservation of mass.

Hence, the equation that conforms with the particle model is 4C + 6O2 ----> 4CO2 + 2O2.

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2 years ago
Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at
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Answer:6.719Litres of Cl2 gas.

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according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

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Going back to the reaction,

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See explanation.

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