The freezing point of the sucrose solution is -0.435°C.
<h3>What is the freezing point of the solution?</h3>
The freezing point of the solution is determined from the freezing point depression formula below:
Kf(H₂O) = 1.86 Cm
m is molality of solution = moles of solute/mass of solvent
moles of sucrose = 8.0/342.3 = 0.0233 moles
m = 0.0233/0.1 = 0.233 molal
ΔT = 0.233 m * 1.86°C/m.
ΔT = 0.435 °C.
Freezing point of sucrose solution = 0°C - 0.435°C
Freezing point of sucrose solution = -0.435°C.
In conclusion, the freezing point of sucrose solution is determined from the freezing point depression.
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Answer:
L=7.0125*10^-5
Explanation:
LET
Length of polyethene chain= L
Molecular weight OF polyethene= MW
Molecular weight of monomer =mw
Monomer length= l
L=(MW/mw) *l
L= (28.05g/mol10^6gmol-1) * 2.5
L=70.125*10^-6
L=7.0125*10^-5
Hence, the length will be L=7.0125*10^-5
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Answer:
B the golgi body and rough/smooth endoplasmic reticulum are able to synthesise vesicles the only difference is that one synthesises vesicles that go from one organelle to another where as the other synthesises vesicles that travel to the outside of the cell.
Explanation:
Answer:
34.59 %
Explanation:
Sodium chloride dissolves in water as follows:
- NaCl(s) → Na⁺(aq) + Cl⁻(aq)
Chloride ions are then precipitated with silver ions:
- Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Now we <u>use the give mass of AgCl to calculate the moles of Cl</u>:
- 0.9805 g AgCl ÷ 143.32 g/mol = 6.841x10⁻³ mol AgCl
mol AgCl = mol Cl
- 6.841x10⁻³ mol Cl * 35.45 g/mol = 0.2425 g Cl
Finally we <u>calculate the mass percent of Cl in the initial mixture</u>:
- 0.2425g / 0.7011g *100% = 34.59 %