By dilution formula.
C1V1=C2V2
molarity of sulphuric acid is 18
18xV1=6x500
V1=6x500/18
V1=166.6ml/500ml of distilled water
FOLLOW THIS FORMULA: N1V1=N2V2
36*V1=6*1000
V1=6000/36
=166ml from conc. H2SO4
To make 6M of an HCl solution, you must first make 12M and divide. Thus, 50 mL of 12M HCl will dilute to 100mL, and will create 6M.
Add 53.3 ml H2SO4 (from 95-97% H2SO4) in 146.7ml..
9.8ml of h2so4 add to 1000 ml water
7n h2so4
In order to determine normality, you have to examine the reaction that produces the sulfuric acid, since normaility is molarity multiplied by proton exchange.
The (N) stands for Normal. 1 Normal is 28 ml of concentrated sulfuric acid added to deionized water for a final volume of 1 Liter (L). 18 N is roughly a 50 percent concentration sulfuric acid.
use the equation: NaVa=NbVb Where N is the normality and V is the volume. NOTE a is the original and b is what you're making. You would sove for Va
how do prepare 0.1 N Oxalic acid
7n h2so4
In order to determine normality, you have to examine the reaction that produces the sulfuric acid, since normaility is molarity multiplied by proton exchange.
The (N) stands for Normal. 1 Normal is 28 ml of concentrated sulfuric acid added to deionized water for a final volume of 1 Liter (L). 18 N is roughly a 50 percent concentration sulfuric acid.
use the equation: NaVa=NbVb Where N is the normality and V is the volume. NOTE a is the original and b is what you're making. You would sove for Va
how do prepare 0.1 N Oxalic acid
Mix 1 part 5,25 N H2SO4 with 4,25 parts water to obtain 1 N H2SO4.
yes
This means that one liter of the solution of sulfuric acid contains 0.2 gram-equivalent mass of sulfuric acid. For this acid, the equivalent mass is one half the molar mass, since each molecule of H2SO4 supplies two hydrogen atoms to neutralize alkaline materials.
0.1 N means you have a hydronium ion concentration of 0.1 mole/L. Since sulfuric acid is a diprotic acid, you get 2 moles of ionizable protons per mole a sulfuric acid. Meaning, you only need a concentration of 0.05 M sulfuric acid to obtain 0.1 N sulfuric acid. [10 L] * [0.1 mole H+/L] * [1 mole H2SO4/2 mole H+] * [98.08 g/mole H2SO4] * [1 mL/1.84 g H2SO4] = 26.65 mL H2SO4 required to make 10 L 0.1 N H2SO4.
The normality of 98 g of sulfuric acid in 500 mL of solution is 4 N
dilute with water
h30 is what comes after h20 ----