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Chapter 3: Q 3-C E (page 60)

(a) How many millilitres of $53.4 (\pm 0.4) wt % NaOH$ with a density of $1.52 (\pm 0.01) g / mL$ will you need to prepare 2.000L of 0.169M NaOH ?
(b) If the uncertainty in delivering NaOH is $\pm 0.01 mL$ , 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

Expert verified

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

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

Step by step solution

01

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 $= (2.000 L) (0.169 M) (\frac{40 g}{1 mol}) = 13.52 gNaOH$

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

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

02

The value of absolute Uncertainty

(b)

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

So, the absolute Uncertainity in the molarity is

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

The uncertainty in delivering is 0.01.

So,

$(0.4 + x + 0.01) = 0.01 x = \pm 0.4 m L$

The absolute Uncertainity is $\pm 0.4 m L$

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Most popular questions from this chapter

We can measure the concentration of HCI solution by reaction with pure sodium carbonate: $2 H^{+} + {Na}_{2} {CO}_{3} \to 2 {Na}^{+} + H_{2} 0 + 105.9884 \pm 0.0007$ required $27.35 \pm 0.04 mL$ of HCI .

(a) Find the formula mass (and its uncertainty) for ${Na}_{2} {CO}_{3}$ .

(b) Find the molarity of the HCI and its absolute uncertainty.

(c) The purity of primary standard ${Na}_{2} {CO}_{3}$ is stated to be 99.95 to 100.5wt % , which means that it can react with $(100.00 \pm 0.05) %$ of the theoretical amount of $H^{+}$ . Recalculate your answer to (b) with this additional uncertainty.

You can calculate the time it will take for an object to fall from the top of a building to the ground if you know the height of the building. If the height has an uncertainty of 1.0% ,what is the uncertainty in time?

Avogadro's number can be computed from the following measured properties of pure crystalline silicon: $^{5}$

(1) atomic mass (obtained from the mass and abundance of each isotope), (2) density of the crystal, (3) size of the unit cell (the smallest repeating unit in the crystal), and (4) number of atoms in the unit cell. For the material that was used, the average atomic mass of Si is role="math" localid="1663313481329" $m_{si} = 28.085 384 2 (35) g / mol$ , where 35 is the uncertainty (standard deviation) in the last two digits. The density is $p = 2.329 031 9 (18) g / {cm}^{3}$ , the size of the cubic unit cell is $c_{0} = 5.431 020 36 (33) \times 10^{- 8} cm$ , and there are 8 atoms per unit cell. Avogadro's number is computed from the equation

$N_{A} = \frac{m_{Si}}{({pc}_{0}^{3}) / 8}$

From the measured properties and their uncertainties, compute Avogadro's number and its uncertainty.

To find the uncertainty of $c_{o n}^{3}$ use the function $y = x^{4}$ in Table 3-1.

What would be the uncertainty in volume delivered if the uncertainty in each reading were 0.03 mL? (Answer: 60.04 mL)

Write each answer with the correct number of significant figures.

(a) $1.0 + 2.1 + 3.4 + 5.8 = 12.3000$

(b) 106.9 - 31.4 = 75.5000

(c) $107.868 (2.113 \times 10^{2}) + (5.623 \times 10^{3}) = 5519.568$

(d) $(26.14 / 37.62) \times 4.38 = 3.043413$

(e $(26.14 / (37.62 \times 10^{8})) \times (4.38 \times 10^{- 2}) = 3.043413 \times 10^{- 10}$

(f) $(26.14 / 3.38) + 4.2 = 4.5999$

(g) $l o g (3.98 \times 10^{4}) = 4.5999$

(h) $10^{- 6.31} = 4.89779 \times 10^{- 7}$

See all solutions

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