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	1/1000^th, i.e. by 10^-3	mmol, mg, ml
µ	micro	1/1000 000^th, 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

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