Answer to Question #187992 in General Chemistry for Ace

Question

A solution of 100.8 g of a non- dissociating solute in 135.0 g of water has a freezing point of -5.16 °C. What is the molar mass of the solute? Freezing point of water= 0 °C Kf=1.86 °C/m

A solution of 1.51 g of a non-dissociating solute in 250.0 g of water is observed to boil at 124.3 °C. Calculate the molar mass of the solute? Boiling point of water=100 °C Kb=0.51 °C/m

Determine the molar mass of a non- dissociating if 20.0 g dissolved in 100.0 mL of solution to give a resulting osmotic pressure of 6.48 atm at 25 °C

Answer 1

1. Mass of solute= 100.8g

Mass of water= 135g= 0.135Kg

∆T_f= 0-(-5.16)= 5.16°C

K_f= 1.86

Let's find the molality of the solute

∆T_f= K_fm

m= 5.16/1.86=2.77molal

But molality= moles of solute/Kg of solvent

Moles of solute= 2.77x0.135=0.374mol

Mole= mass/molar mass

Molar mass= 100.8/0.374= 269.52g/mol

2. Mass of solute[ 1.57g

Mass of water= 250g= 0.25Kg

∆T_b= 124.3-100=24.3°C

K_b= 0.51

Let's find the molality

m=∆T_b/K_b

m= 24.3/0.51=47.65molal

But molality= moles of solute/Kg of solvent

Moles= 47.65x0.25= 11.91mol

Mole= mass/molar mass

Molar mass= 1.57/11.91=0.13g/mol

3. Mass of solute= 20.0g

V= 100ml= 0.1L

P= 6.48atm

T= 298K

PV=nRT

n= 6.48x0.1/0.0821x298

n= 0.026mol

Mole= mass/molar mass

Molar mass= 20.0/0.026= 769.23g/mol