Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
Answer: Motion, providing a burst of power that can move a specific part of the device.
Explanation: Hope this helps
I have no skins for you my brother and my girlfriend and my uncle are going out and have you come on the house for me to do you have to get on your phone
Answer:
Not likely
Explanation:
It would only be likely if the humidity was high enough to rain.
so, it is not likely
Answer:
83.24 mmHg.
Explanation:
- <em>The vapor pressure of the solution (Psolution) = (Xmethanol)(P°methanol).</em>
where, Psolution is the vapor pressure of the solution,
Xmethanol is the mole fraction of methanol,
P°methanol is the pure vapor pressure of methanol.
- We need to calculate the mole fraction of methanol (Xmethanol).
<em>Xmethanol = (n)methanol/(n) total.</em>
where, n methanol is the no. of moles of methanol.
n total is the total no. of moles of methanol and urea.
- We can calculate the no. of moles of both methanol and urea using the relation: n = mass/molar mass.
n of methanol = mass/molar mass = (56.9 g)/(32.04 g/mol) = 1.776 mol.
n of urea = mass/molar mass = (7.38 g )/(60.06 g/mol) = 0.123 mol.
∴ Xmethanol = (n)methanol/(n) total = (1.776 mol)/(1.776 mol + 0.123 mol) = 0.935.
<em>∴ Psolution = (Xmethanol)(P°methanol)</em> = (0.935)(89.0 mmHg) =<em> 83.24 mmHg.</em>
