Answer to Question #306000 in General Chemistry for Joie

Question

B.Calculate the boiling point and freezing point of the following solutions:

1.0.575 molal aqueous solution

2.10 g of C6 H12 O6 dissolved in 259 g of water

3.0.5 mole Br2 dissolved in 507 g chloroform

The boiling point of an aqueous solution is 102.5 degree celcius. What is the freezing point of the solution?

A solution of propylene glycol (antifreeze) in the car’s radiator has a freezing point of -20 0C.What is the molarity of the solution?

Answer 1

The molecular weight 9f glucose C₂H₁₂O₆=6(12)+12(1)+6(16)=180g/mol

The number of moles of glucose=0.625g/180g/mol =0.00347 moles

Mass of water=102.8g

=0.1028kg

The molality of glucose,,,m= 0.00347moles/0.1028 kg= 0.0338 mol/kg.

The depression in the freezing point,∆T_f=k_fm

=1.87K kg/mol×0.0338mol/kg

=0.0632K

Freezing point of water= 273K

The freezing point of the solution=273+0.0632

=273.0632K