Question

must show all work. Observe sig fig rules and include units.

1) In a lab three determinations were made for the freezing point of an unknown liquid. The values obtained were: 25.1°C, 25.9°C and 25.4°C. Calculate the average and Standard Deviation.

2) Compound X was dissolved in water. The freezing point of this solution was determined to be -0.072°C.

a) Calculate the ΔT_f ? (Freezing Point of water is 0.00°C)

b) Calculate the molality of the solution? (K_f for water is 1.858 °C/m)

3) A solution is prepared by dissolving 0.39 grams of an unknown compound in 75.0 grams of water. The molality of this solution was determined to be 0.056 m. Based on this information calculate the following:

a) Moles of the unknown compound (solute)

b) Molecular weight of the unknown compound

Answer 1

Answer: According to the question , Here the average deviation is = 25.1°C+ 25.9°C + 25.4°C. / 3 = 25.46 ⁰C

2] a ] Here the value of the delta T_f is equal to the = 0.072

b] Here we have to used the equation delta T_f = K_f m

hence , molality = 0.072 / 1.858 = 0.0387 m

So, the molality of the solution is 0.0387 m

3] According to the given informations ,

We know that molality = mass /molar mass * volume in kg

molar mass = mass / molality * volume in kg

= 0.39 * 1000 / 0.056 * 75 = 92.857 g /mol

Hence a] Number of moles = mass / molar mass = 0.39 / 92.857 = 0.0042 mol

b] the molecular weight of unknown compound is = 92.857 g /mol .

Thnak you :)