100cm^3 of a sodium hydroxide,NaOH solution contains 4.0g of solid sodium hydroxide,NaOH.40 cm^3 of phosphoric acid,H3PO4 is needed to neutralise 40cm^3 of the alkali.(RAM:H,1;O,16;Na,23;P,31)

(a)Calculate the molarity of the sodium hydroxide,NaOH solution.
(b)Calculate the molarity of the phosphoric acid,H3PO4.

3NaOH + H3PO4 ==> Na3PO4 + 3H2O

mols NaOH = grams/molar mass = 4.0/40 = 0.1 mol.

a) M NaOH = mols/dm3 = 0.1mol/0.1 dm3 = 1M
b) From the equation, 3 mol NaOH = 1 mol H3PO4; therefore, 40 cc (0.04 dm3) of NaOH x 1 M = 0.04 mols NaOH.
mols H3PO4 = 1/3 of that or 0.04/3 = about 0.0133 mols.
Then M H3PO4 = mols H3PO4/dm3 H3PO4.

To calculate the molarity of a solution, we need to use the equation:

Molarity (M) = moles of solute / volume of solution (in liters)

(a) To calculate the molarity of the NaOH solution:
1. Find the number of moles of NaOH in 4.0g:
- The molar mass of NaOH is: (1 x 23) + (1 x 16) + (1 x 1) = 40 g/mol.
- Moles of NaOH = mass / molar mass = 4.0g / 40 g/mol = 0.1 mol.

2. Convert the volume from cm³ to liters:
- 100 cm³ = 100 / 1000 = 0.1 L.

3. Use the formula to calculate the molarity:
- Molarity = 0.1 mol / 0.1 L = 1.0 M.

Therefore, the molarity of the sodium hydroxide solution is 1.0 M.

(b) To calculate the molarity of the phosphoric acid solution:
1. The equation states that 40 cm³ of the NaOH solution is neutralized by 40 cm³ of the phosphoric acid solution. Therefore, the moles of H3PO4 in 40 cm³ will be the same as the moles of NaOH in 40 cm³.

2. Use the number of moles of NaOH calculated earlier (0.1 mol) to find the molarity of H3PO4:
- Molarity = moles of solute / volume of solution (in liters).
- Molarity = 0.1 mol / 0.040 L = 2.5 M.

Therefore, the molarity of the phosphoric acid solution is 2.5 M.