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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

Why do we use quotation marks around the word true in the statement that accuracy refers to how close a measured value is to the "true" value?

Write the formula mass of (a) ${BaF}_{2}$ and (b) $C_{6} H_{4} O_{4}$ with a reasonable number of digits. Use the periodic table inside the cover of this book to find atomic masses.

Write each answer with a reasonable number of figures. Find the absolute and percent relative uncertainty for each answer.

$(a) [12.41 (\pm 0.09) \div 4.16 (\pm 0.01)] \times 7.0682 (\pm 0.0004) = ?$

$(b) [3.26 (\pm 0.10) \times 8.47 (\pm 0.05)] - 0.18 (\pm 0.06) = ? (c) 6.843 (\pm 0.008) \times 10^{4} \div [2.09 (\pm 0.04) - 1.63 (\pm 0.01)] = ? (d) \sqrt{3.24 \pm 0.08} = ? (e) {(3.24 \pm 0.08)}^{4} = ? (f) \log (3.24 \pm 0.08) = ? (g) 10^{3.24 \pm 0.08} = ?$

Compute the molecular mass and its uncertainty for ${NH}_{3}$ . What is the percent relative uncertainty in molecular mass?

Verify the following calculations:

$a) \sqrt{3.1415 (\pm 0.0011) =} 1 . 7724_{3} (\pm 0 . 0003_{3}); b) \log [3.1415 (\pm 0.0011)] = 0 . 4971_{4} (\pm 0 . 0001_{5}); c) antilog [3.1415 (\pm 0.0011)] = 1 . 1447_{0} (\pm 0 . 0003_{5}); d) In [3.1415 (\pm 0.0011)] = 1 . 1447_{0} (\pm 0 . 0003_{5}); e) \log (\frac{\sqrt{0.104 (\pm 0.006)}}{0.0511 (\pm 0.0009)}) = 0 . 80_{0} (\pm 0 . 01_{5})$

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