The Concept Of Equivalent Weight And Molecular Weight

the concept of equivalent weight and molecular weight

         Molarity, which is a unit of concentration that states the number of moles per unit volume (to better understand molarity. Molality, the unit of concentration that states the amount of solute in units of weight. Understanding Normality (N) Completely and this is what we will discuss, that is Normality, let us refer to Understanding Normality (N) Completed With Problem Example. Normality is the unit of concentration that has accounted for the cation or anion contained in a solution and which differs from this Normality, is the calculation of BE or Weight Equivalent. Therefore there is an additional definition for normality. Normality is defined as the number of substances in grams equivalent in one liter of solution with unit N.

The following is the Normality (N) formula:

BE = Mr / The number of H atoms released or received
BE = 98/2 = 49
mass = mol x Mr = 1 x 98 = 98 gr
N = 98/49 x 1 = 2
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2. A total of 5 mL 0.2 N H2SO4 was neutralized with 10 mL of KOH solution (Mr = 56). The KOH mass in 1 liter of KOH solution is
ANSWER:
Mole of acid = M x V
from the previous persoalan it is known that N H2SO4 is 2 times the M. then
H2SO4 0.2N = H2SO4 0.1 M
mole acid = 0.1 x 5 = 0.5 mmol
Remember in the formula neutralization titration
mol equivalent acid = mole base equivalent
n x M x V acid = n x M x V base
To make it easier, you can also use N x V acid = N x V base. the real function of Normality makes calculation easier. then
0.2 x 5 = 1
Base mol = 1 mmol
Mass = mol x Mr = 1 mmol × 56 = 56 mg
1. What is the Normality for 1M H2SO4?
Answer:
H2SO4 => 2H ++ SO42-
1M H2SO4 = 1 mol / L
Formula N = N = \ frac {gram of solute} {BE} \ times \ frac {1000} {mL Solution}
BE = Mr / The number of H atoms released or received
BE = 98/2 = 49
mass = mol x Mr = 1 x 98 = 98 gr
N = 98/49 x 1 = 2
================================================== =====
2. A total of 5 mL 0.2 N H2SO4 was neutralized with 10 mL of KOH solution (Mr = 56). The KOH mass in 1 liter of KOH solution is
ANSWER:
Mole of acid = M x V
from the previous persoalan it is known that N H2SO4 is 2 times the M. then
H2SO4 0.2N = H2SO4 0.1 M
mole acid = 0.1 x 5 = 0.5 mmol
Remember in the formula neutralization titration
mol equivalent acid = mole base equivalent
n x M x V acid = n x M x V base
To make it easier, you can also use N x V acid = N x V base. the real function of Normality makes calculation easier. then
0.2 x 5 = 1
Base mol = 1 mmol
Mass = mol x Mr = 1 mmol × 56 = 56 mg
In 1 L means
56mg × 1000ml / 10ml = 5600mg = 5.6gram