## How do you find the molarity of 85% phosphoric acid?

We simply need the solution density and solute molecular mass (97.994 g·mol−1) to readily calculate the molarity of the commercial 85 [wt]% phosphoric acid. The corresponding density is 1.685 g/cm3 at 25 °C (*). The molarity is: 0.85(g H3PO4 / g aq.

What is the concentration of 85% phosphoric acid?

Therefore, we can say that 1 liter of Phosphoric acid contains 14.615 moles or in other words molarity of 85% (w/w) Phosphoric acid is equal to 14.615 M….Known values.

Known values
Molecular weight of H3PO4 98.00 g/mole
Concentration of Phosphoric acid solution 85% (% by mass, wt/wt)

What is the molar mass of 85% phosphoric acid?

98g/mol
SPECIFICATIONS

Assay % : Min. 85%
Chem Quantity : 1L
Chemical Formula : H3PO4
Density : 1.68
Molecular Weight : 98g/mol

### How do you make 1 M phosphoric acid?

So you need to mix 0.0684 liter (68.4 ml) of concentrated solution to 0.9316 liter (931.6 ml) water to prepare 1 liter of 1 M Phosphoric acid solution.

What is the molarity of phosphoric acid?

Dilutions to Make a 1 Molar Solution

Concentrated Reagent Formula Weight1 Molarity (M)
Phosphoric Acid (H3PO4) 97.995 14.8
Sulfuric Acid (H2SO4) 98.073 18
Ammonium Hydroxide (NH4OH) 35.046 14.5
Sodium Hydroxide (NaOH) 39.997 19.4

What is the molarity of phosphoric acid is 98% H3PO4?

Hence, the molarity of the solution is 17.98 M.

#### How do you make 6 M phosphoric acid?

For example, to make 500 mL of 6M HCl, use 250 mL of concentrated acid and slowly dilute to 500 mL with water….Recipes for Acid Solutions.

Name / Formula / F.W. Concentration Amount/Liter
sp. gr. 1.42 0.1 M 6.3
Phosphoric Acid 6 M 405 mL
H3PO4 3 M 203
F.W. 98.00 1 M 68

What is the molarity of phosphoric acid in a solution Labelled 20.0% phosphoric acid H3PO4 by weight with a density 1.12 g mL?

=11.85 mol/L.

How do you make 10 phosphoric acid?

1. In 100 gram diluted solution you have 1.25 g phosphoric acid.
2. In the 10% solution we have for every gram phosphoric acid 10 gram 10% solution.
3. To end with 1.25 g acid you need 10 x 1.25 = 12.5 g of the 10% solution.
4. This 12.5 gram must be present in 100 gram diluted solution, so you must add 100 – 12.5 = 82.5 gram water.