Answer:
474.64mmHg
Explanation:
From the question given, we obtained the following:
V1 = 10L
P1 = 655mmHg
V2 = 13.8L
P2 =?
P1V1 = P2V2
10 x 655 = P2 x 13.8
Divide both side by the coefficient P2 i.e 13.8
P2 = (10 x 655) / 13.8
P2 = 474.64mmHg
Therefore, the new pressure will be 474.64mmHg
Answer:
1.62
Explanation:
From the given information:
number of moles of benzamide 
= 0.58 mole
The molality = 
= 0.6837
Using the formula:
where;
dT = freezing point = 27
l = Van't Hoff factor = 1
kf = freezing constant of the solvent
∴
2.7 °C = 1 × kf × 0.6837 m
kf = 2.7 °C/ 0.6837m
kf = 3.949 °C/m
number of moles of NH4Cl = 
= 1.316 mol
The molality = 
= 1.5484
Thus;
the above kf value is used in determining the Van't Hoff factor for NH4Cl
i.e.
9.9 = l × 3.949 × 1.5484 m
l = 1.62
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
Explanation:
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