Answer:
Here is your Answer.......
Explanation:
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This 273 K and 101.3 kPa is Standart conditions, so 22.4 L is 1 mol, 1 mol anything contains <span>6.023 x 10^23,
AnswerA</span>
Answer:
B. all the above no wrong answer
0.37 kilogramme is the same as 370 grams
<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
