Larutan Elektrolit Non elektrolit

Non Electrolyte Solution

GROUP 1BERLY DWIKARYANI – ERLINDA

PURNAMA – META TANJUNG – N DEA NAOMI – RICO JULIAN - RISKI

YUDATAMA

Non-Electrolyte Solution

Non-electrolyte solution is a solution that can not conduct electricity, it is because the solution can not produce ions (not to ions).

• Solution of urea• A solution of sucrose• A solution of glucose• A solution of alcohol and other

Properties Of Nonelectrolyte

Solution• Can not conduct electricity• Not a process of ionization• When electrolyte test light is not on

and no gas bubbles

The degree of ionization

More or less The substances molecules that ionized are expressed in degrees of ionization. The degree of ionization (α) is the ratio of the number of molecules of a substance that breaks down to the number of molecules of substances initially.

α = 1, a strong electrolyte0 <α <1, a weak electrolyteα = 0, non-electrolyte

Problem 13.16 grams of a non-electrolyte

compounds dissolved in 75 ml of water. The density of the solution is 0.779 g/mol which consists of 42.9% C; 2.4% H; 16.6% N; 38.1% O. If we know the boiling point increase was at 6.5oC and Kb = 20.2oC/m, determine the molecular formula of the compound?

Solution 1Mass of solute = 3,16 gramsVolume of solvent = 75 mlρ solvent = 0,779 gr/ml

Mass of solvent = 75ml x 0,779 gr/mol= 58,425 grams

ΔTb = 6,5oC

Kb = 20,2oC/m

C H N O

42,9% 2,4% 16,6% 38,1%

42,9 grams 2,4 grams 16,6 grams 38,1 grams

42,9/12 = 3,57

2,4/1 = 2,4

16,6/14 = 1,2

38,1/16 = 2,4

3 2 1 2

So we can get the empirical formula of the compound isC3H2NO2

ΔTb = m . Kb

6,5 =

6,5 =

6,5 =

379,76 Mr = 63832 Mr = 168,085

(C3H2NO2)n = Mr

(C3H2NO2)n = 168,085

36n + 2n + 14n + 32n = 168,085 84n = 168,085

n = 2So,(C3H2NO2)n = (C3H2NO2)2

= C6H4N2O4

Problem 224 grams of a non-electrolyte

substances dissolved in water until the volume is up to 250 ml and has an osmotic pressure of 32.8 atm at 27oC of temperature. If R = 0,082 L atm/mol K, relative molecular mass of the substance is ...

Solution 2Mass of solute = 24 gramsVolume of solvent = 250 mlπ = 32,8 atm

T = 27oC + 273 = 300 K

R = 0,082 L atm/mol K

Mr...?ρ solvent= m solvent/v solvent1 gr/ml = m/250 ml

Mass of solvent = 250 grams

π = M . R . T

π =

32,8 =

32,8 =

Mr solute = = 72 gr/mol

Problem 3How many grams of KCL should be dissolved into 100 g of water so that water boils at 101oC, if the degree of ionization of KCl 0.9 and the price of water 0.52OC/m ?

Solution 3P = 100 grams

Tb = 101oC

Kb = 0,52oC/m

α = 0,9

Mass of KCl…?

ΔTb = m . Kb . I

101-100 =

1 =

1 =

Mass of solute = = 7,54 grams

THE END