In a experiment of Decrease of the freezing point and determination of molar mas

Question

In a experiment of Decrease of the freezing point and determination of molar mass of a non volatil solute, I got 15ml of cyclohexane (C6H12 ) weighting 11.11g and I put it on Ice to freeze it initial temp. was 22C and I take 15 readings of temp 12C,8C,8C,7C,7C,7C,6C,5C,4C,3C,-1C,-2C,-3C,-3C,-3C and the it has to be defrezze on our hands and the final temp. it defreeze was 19C , after these I put inside the solution already defreeze at this time 22C , 0.15 g of Naphthalene ( C10H8 ) and put this back again on Ice and readings now are 10C,5C,5C,4C,4C,3C,2C,2C,2C,2C,-1C,-1C,-1C and then defreeze it again until final temp 17C so, Show the calculation of the decrease in the freezing point (Tf) and Show the calculation of molar mass of the solid ( Solute ) that is Naphthalene C10H8 , Thank You is for tomorrow so dont be busy

Explanation / Answer

Weight of cylcohexane taken = 11.11 g = (11.11 g)*(1 kg/1000 g) = 0.01111 kg.

The freezing point of pure cyclohexane is 19C; the freezing point of cyclohexane mixed with naphthalene is 17C. Therefore, the depression in freezing point due to addition of naphthalene is Tf = (freezing point of pure cyclohexane) – (freezing point of cyclohexane + naphthalene) = (19C) – (17C) = 2C (ans).

Mass of naphthalene taken = 0.15 g.

Use the relation Tf = Km.m where Km = molal freezing point depression constant.

To determine the molar mass of naphthalene, we will need to know KM. You haven’t included Km in your question. Please include that so that the next part can be answered.

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