<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>
the molar mass of the element is 81.36 g/mol
<u><em>calculation</em></u>
step 1 : multiply each %abundance of the isotope by its mass number
that is 79.95 x 29.9 =2391
81.95 x 70.1 = 5745
Step 2: add them together
2390.5 + 5744.7 =8136
Step 3: divide by 100
= 8136/100 = 81.36 g/mol
Answer:
A. the prevailing westerlies
Ah, this is actually a fairly simple chemistry question.
Recall that at STP, one mole is equivalent to 22.4 L. Knowing this, all you have to do is divide 500 L by 22.4 L to find the number of moles:
500/22.4 = 22.3 mol
-T.B.
