CHEMISTRY 31
Fall, 2017 - Dixon
Homework Set 1.2 Solutions
Ch. 1: 37 (carried forward from set 1.1)
37. How many grams
of 0.491 wt % aqueous HF are required to provide a
50% excess to react with 25.0 mL of 0.0236 M Th^{4+} by the reaction Th^{4+}
+ 4F^{-} => ThF_{4} (s)?
mmoles Th^{4+} = (23.6 mmol/L)*(25.0 mL)*(10^{-3} L/mL) = 0.590 mmol
mmol F^{-} required = 0.59*(4 mol F^{-}/1 mol Th^{4+}) = 2.36 mmol
mmol F^{-} required for 50% excess = 1.5*(2.36 mmol)
= 3.54 mmol
g HF sol'n needed = (20.0 mg HF/mmol HF)*(3.54 mmol HF)*(10^{-3}g/mg)/(0.00491
mgHF/mg sol'n) = 14.4 g.
Ch. 3: 1, 2, 5a-d, 11, 14, 16a-f
1. How many significant
figures are there in the following numbers?
a) 1.9030 – 5 (all 0s significant) b) 0.03910 – 4 (all 0s to right of non-0 digits significant)
c) 1.40 x 10^{4} – 3 (all 0s significant)
2. Round each number as indicated:
a)
1.2367 to 4 significant figures – 1.237 b) 1.2384 to 4 significant figures – 1.238
c)
0.1352 to 3 significant figures – 0.135 d) 2.051 to 2 significant figures – 2.1
e) 2.0050 to 3 significant
figures – 2.00 (when last digit to
round is exactly 5, or 5000..., you round up or down depending on whether the
next digit to the left is even or odd)
5. Write each answer
with the correct number of digits.
a)
1.021 + 2.69 =
3.711 -> 3.71 (hundredths)
b)
12.3 - 1.63 =
10.67 -> 10.7 (tenths)
c)
4.34
x 9.2 = 39.928 -> 40 or 40. or 4.0 x 10 (2 significant digits)
d) 0.0602/(2.113 x 10^{4})
= 2.84903 x 10^{-6} -> 2.85 x 10^{-6} (3 sig. digits)
11. Suppose that in a
gravimetric analysis, you forget to dry the filter crucibles before collecting
precipitate. After filtering the
product, you dry the product and crucible thoroughly before weighing them. Is the error in the mass of product that you
report a systematic or a random error? The error is a systematic error.
Is it always high or always low?
The measured product mass is always lower than the true
mass. The initial mass was too high, and
since this is subtracted, the product mass will be too low.
14. Rewrite the number
3.12356 (+0.16789%) in the forms
a) number
(+ absolute uncertainty) and b)
number (+ percent relative) uncertainty with an appropriate number of
digits
a) absolute uncertainty = (0.16789/100)3.12356 = 0.005244 (= 0.005 since only 1
sig fig for uncertainty); number (+ absolute uncertainty) = 3.124 + 0.005
b) 3.124
+ 0.2% (the number of sig figs in the number remains the same)
16. Find the absolute and
percent relative uncertainty and express each answer with a reasonable number
of significant figures.
a)
9.23(+0.03) + 4.21 (+0.02) - 3.26 (+0.06)
= ?
value = 10.18
since only +/-, uncertainty = [(0.03)^{2} + (0.02)^{2} +
(0.06)^{2}]^{0.5} uncertainty = 0.07
% uncertainty = 0.07*100/10.18 = 0.7%
answer =
10.18 (+0.07)
b)
91.3 (+1.0) * 40.3 (+0.2)/21.1 (+0.2)
= ?
value = 174.3787
since only * and /, % rel. uncertainty = [(1.0*100/91.3)^{2} +
(0.2*100/40.3)^{2} + (0.2*100/21.1)^{2}]^{0.5}
% uncertainty = 1.5%
= 2%
uncertainty = 1.5*174.3787/100= 2.67 = 3
answer =
174 (+3)
c)
[4.97 (+0.05) - 1.86 (+0.01)]/21.1 (+0.2)
= ?
Subtraction part first: 4.97 - 1.86 = 3.11; uncertainty = [(0.05)^{2} + (0.01)^{2}]^{0.5} = 0.051
Division part: 3.11/21.1 =
0.1474
rel. unc. = [(0.051/3.11)^{2} +
(0.2/21.1)^{2}]^{0.5} = 0.0189 (%
rel unc. = 2%)
uncertainty = 0.189*0.1474 = 0.0028
answer =
0.147 (+0.003)
d) 2.0164
(+0.0008) + 1.233 (+0.002) + 4.61 (+0.01) = ?
value = 7.8594
since only +/-, uncertainty = [(0.0008)^{2} + (0.002)^{2} +
(0.01)^{2}]^{0.5}
uncertainty =
0.0102
% uncertainty = 0.0102*100/7.8594 = 0.1%
answer = 7.86
(+0.01)
e)
2.0164 (+0.0008) x 10^{3} + 1.233 (+0.002)
x 10^{2} + 4.61 (+0.01) x 10^{1} = ?
value = 2185.8
since only +/-, uncertainty = [(0.8)^{2} + (0.2)^{2} + (0.1)^{2}]^{0.5}
uncertainty =
0.831
% uncertainty = 0.831*100/2186 = 0.04%
answer = 2185.8
(+0.8)
f) [3.14 + 0.05]^{1/3}
= ?
value = 1.464
rel. unc. = (1/3)(0.05/3.14)
= 0.0053
unc. = (0.0053)(1.464) = 0.0078
% uncertainty =0.5%
Answer = 1.464+0.008
14
[postpone to set 1.3]
1.
What is the relation between the standard deviation and the precision of
a procedure? What is the relation
between standard deviation and accuracy?
The standard deviation
is directly related to precision of a procedure where low standard deviations
are indicative of a high level of precision.
There is not a
necessary relation between standard deviation and accuracy except that a high
level of accuracy requires low standard deviations. One can also have low standard deviations and
poor accuracy if systematic errors exist.
3.
The ratio of the number of atoms of the isotopes ^{69}Ga and ^{71}Ga
in eight samples from different sources was measured in an effort to understand
differences in reported values of the atomic mass of gallium:
Sample |
^{69}Ga/^{71}Ga |
Sample |
^{69}Ga/^{71}Ga |
1 |
1.52660 |
5 |
1.52894 |
2 |
1.52974 |
6 |
1.52804 |
3 |
1.52592 |
7 |
1.52684 |
4 |
1.52731 |
8 |
1.52793 |
Find the a) mean
Mean = 1.52767 (used calculator functions)
b) standard deviation
0.00126 (used calculator functions)
c) standard deviation of the mean
standard deviation of the mean = S/(n)^{0.5}
= 0.00126/(8)^{0.5} = 0.00045
5. a) Calculate the
fraction of bulbs in Figure 4-1 expected to have a lifetime greater than
1005.3 hours.
First, calculate the average and standard
deviation: average = 845.2 h
standard deviation = 94.2
h
This can be determined first by calculating how far is 1005.3 past the
average in terms of the standard deviations = z = (1005.3 - 845.2)/94.2 =
1.700.
Next, the area under the Gaussian curve between zero
and 1005.3 h can be found as the area value in Table 4-1 for z = 1.7 and is
0.4554. The fraction greater than 1005.3
is equal to 0.5 – the table area:
Fraction = 0.500 - 0.4554 = 0.045
b)
Calculate the fraction expected to have a lifetime between 798.1 and 901.7 h.
z(798.1) = (798.1 -
845.2)/94.2 = -0.500
z(901.7) = (901.7 -
845.2)/94.2 = 0.600
Area(z = -0.500) =
Area(z = +0.500) = 0.1915
Area(0.600) = 0.2258
Area(798.1 < x <
901.7) = Area(0.500) + Area(0.600).
Using Table 4-1 and linear interpolation [used to get values between
columns or rows in Table – this won’t be on an exam],
Fraction = 0.1915 + 0.2258 = 0.417
10.
What fraction of vertical bars in Figure 4-5a is
expected to include the population mean (10,000) if many experiments are
carried out?
50% of the samples should be within the 50% confidence interval of the
true values.
Why are the 90% confidence
interval bars longer than the 50% bars in Figure 4-5?
They are longer because they have a greater chance of including the
true value.
12.
The percentage of an additive in
gasoline was measured six times with the following results: 0.13, 0.12, 0.16, 0.17, 0.20, 0.11%. Find the 90%
and 99% confidence intervals for the percentage of the additive.
mean value = 0.148%;
standard deviation = 0.034% (absolute value)
90 % Confidence Interval: t (n-1 = 5) = 2.015
confidence interval = mean±t·s/(n)^{0.5} = 0.148±2.015·0.034/(6)^{0.5}
= 0.15±0.03%
99% confidence Interval: t (n-1 = 5) = 4.032
confidence interval = mean±t·s/(n)^{0.5} =
0.148±4.032·0.034/(6)^{0.5} = 0.15±0.06%