Question

(a) How many millilitres of $\mathbf{53}\mathbf{.}\mathbf{4}\left(\pm 0.4\right)\mathbf{wt}\mathbf{\%}\mathbf{NaOH}$with a density of $\mathbf{1}\mathbf{.}\mathbf{52}\left(\pm 0.01\right)\mathbf{g}\mathbf{/}\mathbf{mL}$will you need to prepare 2.000L of 0.169M NaOH ?

(b) If the uncertainty in delivering NaOH is $\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{mL}$, calculate the absolute uncertainty in the molarity (0.169M). Assume there is negligible uncertainty in the formula mass of NaOH and in the final volume (2.000L).

Short answer

(a) 16.65 mL of NaOH is required to prepare 2.000L of 0.169M NaOH.

(b) The absolute Uncertainity is$\pm 0.4\mathrm{mL}$

Step-by-step solution

The requirement of NaOH

(a)

For the preparation of 2.000L, 0.169M NaOH, from 53.4 wt% of NaOH.

The mass of the NaOH which are required$=\left(2.000\mathrm{L}\right)\left(0.169\mathrm{M}\right)\left(\frac{40\mathrm{g}}{1\mathrm{mol}}\right)=13.52\mathrm{gNaOH}$

So, the quantity of solution is$=\frac{100}{53.4}\times 13.52\mathrm{g}\times \frac{1\mathrm{mL}}{1.52\mathrm{g}}=16.65\mathrm{mL}$

16.65 mL of NaOH is required to prepare 2.000L of 0.169M NaOH.

The value of absolute Uncertainty

(b)

The mass of NaOH is$=\left(2.000\mathrm{L}\right)\left(0.169\mathrm{M}\pm \mathrm{x}\right)\left(\frac{40\mathrm{g}}{1\mathrm{mol}}\right)=\left(13.52g\pm x\right)\mathrm{NaOH}$

So, the absolute Uncertainity in the molarity is

$=\left(\frac{100}{53.4\pm 0.4}\times \left(13.52\mathrm{g}\pm \mathrm{x}\right)\times \frac{1\mathrm{mL}}{1.52\mathrm{g}\pm 0.01}\right)=16.65\mathrm{mL}\pm \left(0.4+x+0.01\right)$

The uncertainty in delivering is 0.01.

So,

$$\begin{gathered} \left(0.4+x+0.01\right)=0.01 \\ x=\pm 0.4mL \end{gathered}$$

The absolute Uncertainity is $\pm 0.4mL$