1) The boiling point of water/ H 2 O is 100.00 °C at 1 atmosphere. If 14.23 gram

Question

1) The boiling point of water/H2O is 100.00 °C at 1 atmosphere. If 14.23 grams of aluminum acetate, (204.1 g/mol), are dissolved in 293.7 grams of water ...The molality of the solution is  m. The boiling point of the solution is  °C.

2) The boiling point of water is 100.0°C at 1 atmosphere. (Kb(water) = 0.512°C/m)   How many grams of barium bromide (297.1 g/mol), must be dissolved in 249.0 grams of water to raise the boiling point by 0.450°C ? ... g barium bromide.

3) The freezing point of water is 0.00°C at 1 atmosphere. If 11.04 grams of aluminum chloride, (133.3 g/mol), are dissolved in 196.4 grams of water. The molality of the solution is  m. The freezing point of the solution is  °C.

4) An aqueous solution of aluminum acetate has a concentration of 0.480 molal. The percent by mass of aluminum acetate in the solution is  %.

5) An aqueous solution is 13.0% by mass ethanol, CH3CH2OH, and has a density of 0.978 g/mL. The molality of ethanol in the solution is  m.

Explanation / Answer

1)

Lets calculate molality first

mass,m = 14.23 g

number of mol,

n = mass/molar mass

= 14.23/204.1

= 0.0697 mol

massof solvent ,m = 293.7 g = 0.2937 Kg

Molality,

m = number of mol / mass of solvent in Kg

= 0.0697/0.2937

= 0.237 M

lets now calculate Tb

Tb = Kb*m

= 0.512*0.237

= 0.122 oC

This is increase in boiling point

freezing point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.122

= 100.122 oC

Answer: 100.122 oC

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