Freelancer Basic Numeracy Level 1 Answers 2016
Basic Numeracy Level 1 Answers of Freelancer 2016
Freelancer Basic Numeracy Level 1 Answers
Question 1
How many mg (milligrams) are in 0.8 g?
There are 1000 mg in 1 g. So in 0.8 g there are 1000 x 0.8 = 800 mg
Question 2
How many ml (milliltres) are in 2,500 µl (microlitres)?
1000 µl = 1 ml, so 2,500 µl = 2,500/1000 ml = 2.5 ml
Question 3
How many mg (milligrams) are in 5250 µg (micrograms)?
1000 µg = 1 mg, so 5,250 µg = 5,250/1000 mg = 5.25 mg
Question 4
How many ng (nanograms) are in 5250 µg (micrograms)?
1 µg = 1000 ng, so 5,250 µg = 5,250 x 1000 ng = 5,250,000 ng
Question 5
Which of the following is a CONCENTRATION and not an AMOUNT?
2ml is a volume (amount) and 10g and 100µg are masses (amount); 100mM = 100 mmol/l and is a concentration.
Question 6
Which of the following is an AMOUNT and not a CONCENTRATION?
10%(w/v), 35 µM (micromolar) and 20 mmol/l (millimolar) are all concentrations. 8 mmol (millimoles) is an amount.
Question 7
The molecular weight of the amino acid alanine is 89. How many mg are in 30 mmol (millimoles) of alanine?
1 mol alanine = 89 g, so 1 mmol = 89 mg, so 30 mmol = 30 x 89 mg = 2670 mg
Question 8
The molecular weight of the amino acid alanine is 89. How many g are in 45 µmol (micromoles) of alanine?
1 mol alanine = 89 g, so 1 µmol = 89 µg = 89 x 10-6 g = 0.000089 g, so 45 µmol = 0.000089 x 45 g = 0.004005 g.
Question 9
How many µmol (micromoles) are dissolved in 1 l of a 30 µM (micromolar) solution?
30 µM = 30 µmol/l, so in 1 litre there are 30 µmol.
Question 10
How many µmoles are in 1 ml of a 1 M solution?
1 litre of a 1 M solution contains 1 mol, so 1 ml contains 1 mmol = 1000 µmol.
Question 11
A solution was made by dissolving 7.687g of substance X in 720ml of water. Calculate the concentration of X in g/litre.
If 720 ml contains 7.687 g, then 1000 ml (1 l) contains 7.687 x 1000/720 g = 10.68g. Concentration is 10.68 g/l.
Question 12
If
the molecular weight of substance X is 212 what is the molarity of the
previous solution, given that it had a concentration of 10.68 g/l?
212 g/l = 1 M, so 10.68 g/l = (1/212) x 10.68 M = 0.05 M
Question 13
The molecular weight of the amino acid alanine is 89. How many mg are in 250 ml of a 10 µM solution of alanine?
1
l of a 1M solution contains 89 g, so 250 ml of a 1 M solution contains
89 x 250/1000 = 22.25 g, so 250 ml of a 1 µM solution contains 22.25 µg,
so 250 ml of a 10 µM solution contains 222.5 µg = 0.2225 mg.
Question 14
The molecular weight of the amino acid alanine is 89. How many µg are in 200 ml of a 0.02% (w/v) solution of alanine?
A 0.02% (w/v) solution contains 0.02 g/100ml, so 200 ml contains 0.04 g = 0.04 x 106 µg = 40,000 µg. The molecular weight is not required for the calculation.
Question 15
The molecular weight of the amino acid alanine is 89. How many g are in 1.0 kg of a 50 ppm solid mixture of alanine in salt ?
50 ppm (parts per million) = 50 g per 106 g = 50 g per 103 kg = 50 x 10-3 g per kg = 0.05 g/kg. The molecular weight is not required.
Question 16
The molecular weight of glucose is 180. Express a blood glucose concentration of 80 mg per 100 ml as a molarity.
80 mg/100ml = 800 mg/l = 0.8 g/l. 180 g/l = 1 M, so 0.8 g/l = 0.8/180 M = 0.0044 M = 4.4 mM.
Question 17
The molecular weight of haemoglobin is 64,000. What is the concentration of a 5 µM solution of haemoglobin expressed in mg/ml?
1M Hb = 64,000 g/l, so 1 µM Hb = 64,000 µg/l, so 5 µM Hb = 320,000 µg/l = 320 mg/l = 0.32 mg/ml.
Question 18
The
molecular weight of glucose is 180. 500 ml of a 10% (w/v) solution is
made and 100 ml of this is removed. What is the molar concentration of
glucose in the remaining 400 ml?
10% (w/v) glucose contains 10 g/100 ml = 100 g/l. 180 g/l is 1 M, so 100 g/l = 100/180 M = 0.555M. The fact that 500 ml were made and 100 ml removed does not affect the concentration. It is the same i.e. 10% (w/v), or 0.555 M, in the 500ml, the 400 ml and the 100ml.
Question 19
The
labelling on a tube of toothpaste states that it contains 0.1% (w/w)
sodium fluoride (NaF). The content is also expressed as "x" ppm fluoride
ion. What is "x"? (The molecular weight of NaF is 42 and the atomic
weight of F is 19).
0.1% (w/w) is 0.1 g NaF per 100g toothpaste, so 106 g toothpaste would contain 0.1 x 106/100 g = 1000 g NaF. This is 1000 ppm NaF, so fluoride ion (F-) is 1000 x 19/42 = 450 ppm.
Question 20
The
molecular weight of urea is 60. 5 g were dissolved in 50 ml water. 10
ml of this solution was added to 50 ml water. What was the molarity of
this final diluted solution?
5
g in 50 ml = 100g/l. 60 g/l = 1M, so 100 g/l = 100/60 M = 1.67 M. 10 ml
of 1.67 M urea added to 50 ml water, so final volume = 60 ml. This is a
10/60 or 1 in 6 dilution, so molarity of diluted solution = 1.67/6 = 0.278 M.
| Prefix |
Name
|
Which modifies an amount by
|
Examples
|
m
|
milli
|
1/1000th, i.e. by 10-3
|
mmol, mg, ml
|
µ
|
micro
|
1/1000 000th, i.e. by 10-6
|
µmol, µg, µl
|
n
|
nano
|
by 10-9
|
nmol, ng
|
p
f
k
M
|
pico
femto
kilo
mega
|
by 10-12
by 10-15
by 1000 times, i.e. by 103
by 106
|
pmol, pg
fmol, fg
kg
MPa
|
Comments
Post a Comment