Question

you need 16.66ml (+-0.01) of 53.4 (+-0.4)wt% of NaOH with a density of 1.52 (+-0.01)g/mL to prepare 2.00L of 0.169M of NaOH. What is the uncertainty in the molarity of NaOH. FM of NaOH=39.9971 g/mol. Neglect the uncertainty in the final volume and FM.

Answer 1

The absolute uncertainty is approximately 1.69 × 10⁻³

Step-by-step explanation:

The volume needed for NaOH needed to make the solution = 16.66 ml

The wt% of the added NaOH = 53.4 wt%

The volume of the NaOH to be prepared = 2.00 L

The concentration of the NaOH to be prepared = 0.169 M

The molar mass of NaOH = 39.997 g/mol

Therefore, 100 g of sample contains 53.4 g of NaOH

The mass of the sample = 16.66 × 1.52 = 25.3232 g

The mass of NaOH in the sample = 0.534 × 25.3232 = 13.5225888 g ≈ 13.52 g

Therefore;

The number of moles of NaOH = 13.52/39.9971 = 0.3381 moles

Therefore, we have 0.3381 moles in 2.00L solution, which gives;

The number of moles per liter = 0.3881/2 = 0.169045 moles/liter

The molarity ≈ 0.169 M

The absolute uncertainty, u(c) is given as follows;

`u(c) = \sqrt{ \left ((0.01)/(16.66) \right )^2 + \left ( (0.4)/(53.4) \right )^2 + \left ( (0.01)/(1.52) \right )^2 } * 0.169 \approx 1.69 * 10^(-3)`

The absolute uncertainty, u(c) ≈ 1.69 × 10⁻³.