If you have a increased number of rod cells
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Answer:
ethylene glycol (molar mass = 62.07 g/mol).
Explanation:
- We can solve this problem using the relation:
<em>ΔTf = Kf.m,</em>
ΔTf is the depression in the freezing point of tert-Butyl alcohol (ΔTf = freezing point of pure solvent - freezing point in presence of unknown liquid = 25.5°C - 15.3°C = 10.2°C).
Kf is the molal freezing point constant of tert-Butyl alcohol (Kf = 9.1 °C/m).
m is the molality of unknown liquid.
∵ ΔTf = Kf.m
<em>∴ m = ΔTf/Kf </em>= (10.2°C)/(9.1 °C/m) = <em>1.121 m.</em>
- We need to calculate the molar mass of the unknown liquid:
Molality (m) is the no. of moles of solute in 1.0 kg of solvent.
∴ m = (mass/molar mass) of unknown liquid/(mass of tert-Butyl alcohol (kg))
m = 1.121 m, mass of unknown liquid = 0.807 g, mass of tert-Butyl alcohol = 11.6 g = 0.0116 kg.
<em>∴ molar mass of unknown liquid = (mass of unknown liquid)/(m)(mass of tert-Butyl alcohol (kg)) </em>= (0.807 g)/(1.121 m)(0.0116 kg) = <em>62.06 g/mol.</em>
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- So, the unknown liquid is:
<em>ethylene glycol (molar mass = 62.07 g/mol).</em>
Answer:
I dont know how to take a picture so I will just tell you the answer. The answer is no.
Explanation:
Hope this helps. :) :D
The answer in here is d)48 lit. We know this when we do the following procedure:
<span>kerosene=18%
so petrol =82%
after
kerosene =15%
petrol=85%
so 3% =8
lit
so total kerosene in the mixture =(8x18/3) =48 lit
</span>Hope this is useful
