Data:
<span>Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
Vapor pressure of water at 343 k: 233.7 torr.
Quesiton: Vapor pressure of water
Solution:
Raoult's Law: </span><span><span>The vapour
pressure of a solution of a non-volatile solute is equal to the vapour
pressure of the pure solvent at that temperature multiplied by its mole
fraction.
Formula: p = Xsolvent * P pure solvent
X solvent = moles solvent / moles of solution
molar mass of H2O = 2*1.0g/mol + 16.0 g/mol = 18.0 g/mol
moles of solvent = 135 g of water / 18.0 g/mol = 7.50 mol
molar mass of C3H8O3 = 3*12.0 g/mol + 8*1 g/mol + 3*16g/mol = 92 g/mol
moles of solute = 28.5 g / 92.0 g/mol = 0.310 mol
moles of solution = moles of solute + moles of solvent = 7.50mol + 0.310mol = 7.810 mol
Xsolvent = 7.50mol / 7.81mol = 0.960
p = 233.7 torr * 0.960 = 224.4 torr
Answer: 224.4 torr
</span> </span>
Answer:
401135 kJ
Explanation:
From the balanced quation,
(q/n) = CΔE
Molar heat of combustion = 7.85kJk × (303.81-298.70)k
= 7.85kj × 5.11
= 40.1135kj
The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.
Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be 
g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.
