Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample = 
= 165°C
Depression in freezing point = 
Depression in freezing point is also given by formula:
= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have: = 40°C kg/mol
i = 1 ( organic compounds)
The molality of isoborneol in camphor is 0.53 mol/kg.
Answer:
work - unchanged
plane-tilted
constant- A surface
inclined- Effort exerted over a distance
Answer is: mass of <span>potassium bromide is 4.71 grams.
V(KBr) = 25.4 mL </span>÷ 1000 mL/L = 0.0254 L, volume of solution.
c(KBr) = 1.56 mol/L.
n(KBr) = c(KBr) · V(KBr).
n(KBr) = 1.56 mol/L 0.054 L.
n(KBr) = 0.0396 mol, amount of substance.
m(KBr) = n(KBr) · M(KBr).
m(KBr) = 0.0396 mol · 119 g/mol.
m(KBr) = 4.71 g.
M - molar mass.
Answer:
Explanation:
Hello!
In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:
Then, the liters of solution:
Finally, the molarity turns out:
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