Finally we reached the last Question of the 3rd year practical exam sample paper:) I hope you did enjoy the problem solutions provided and that you feel a little bit more confident about the exam. Remember to believe in your skills. That will defenitely help you to go through the exam questions smoothly:)

Aha I almost forgot, it is getting better beacause there is more to come:) In the near future you will find here (in the different section called 3rd year Biochemistry) a sample paper of the lecture exam with the solutions. Can you believe it:) Ok let's finish first the practical paper:)

All let's rollin' baybe! Question 6 Part a) This question is very simple. First two parts are about mixing stuff:) You are asked to explain how to prepare 250ml of 25% (w/v) ammonium chloride solution. Remember that:

(w/v) stands for weight to volume

(v/v) stands for volume to volume

Usually buffers or solutions are prepared with water. If you remember the density of the water is roughly 1.0kg/L = 1g/ml so it means that 250ml of water weights 250g. So mass of our solution should be around 250g. We have

mass of the ammonium chloride = 250g x 25% = 250g x 0.25 =

**62.5g**So to prepare our solution we need 62.5g of the ammonium chloride and fill it up with the water to 250ml to get in total 250g of the solution. So we need to add about 250g - 62.5g = 187.5g of water =

**187.5ml**of water:)In the second part you are asked to calculate what is the concentration of the solution c2 if you resuspend 60g of ammonium chloride in the 200ml of water. So:

200ml water = 200g of water = Solvent Mass

60g of ammonium chloride = Substance Mass

%Concentration C2 = (Substance Mass / Solution Mass) x 100%

where

Solution Mass = Substance Mass + Solvent Mass

We have:

%Concentration C2 = 60g/(60g + 200g) x 100% =

**23%**Part b)

**You are asked about the same stuff here, but the author of the question wanted to complicate things a little bit more (but we will get it anyway:). So we need to prepare 250ml of the 0.4mol/L = 0.4M sodium chloride (58.4g/mol) solution which is supplemented with the 0.2% sodium azide (65.01g/mol). First we will deal with the sodium chloride. We have:**Concentration of NaCl = 0.4mol/L = 0.4M

Finval Volume = 250ml = 0.25L

**Remember this simple equation:****Molar Concentration = n (number of moles) / v (volume)**number moles of the NaCl required = 0.25L x 0.4mol/L = 0.1mol

We know that 1mol of the sodium chloride weights 58.4g, so how much 0.1mol weights:)

1mol - 58.4g

0.1mol - xg

x = (58.4g x 0.1mol) / mol =

**5.84g mass of the NaCl required**

We calculate mass of the the sodium azide in the analogous way to the part a) So,

Mass of the sodium azide = 250g x 0.2% = 250g x 0.002 = 0.5g

And again we fill it up to 250ml with water.

In the second part you are asked to calculate the concentration of sodium chloride and sodium azide if you dilute prepared solution to 1L. This is easy:) First we calculate the dilution factor:

dilution factor = final volume / starting colume = 1L / 0.25L =

**4** So we know that we have diluted our solution 4 times. So:

final concentration of sodium chloride = starting concentration / 4 = 0.4M / 4 =

**0.1M**final concentration of sodium azide = starting concentration / 4 = 0.2% / 4 =

**0.05%**Part c) Here you need to explain what accuracy and standard deviation means.

Accuracy - (simply saying) it is clossenest of the measurment to its actual (real) value.

Standard deviation - (simply saying:) tells you how thightly different measurments are clustered around the mean in a set of the data.

Part d) In this part you need to actually calculate the standard deviation of different experiments (A1 - A9) using adsorbance as your sample readout. We know that:

where:

SD - standard deviation

Xi - value of single measurment

X with the dash = mean of all measurments in experiment

n - number of samples

The other creepy stuff in that formula means that you have to sum up (greek sigma letter) square of the difference between sample and the mean for of all the samples. To get our SD we prepare Table:

That is it:) After the SD calcilation is done you need to comment on the precision of each experiment. So first we will look what the precision is:)

Precision - which might be called reproducubility or repeatabillity as well is the degree to which repeated measurments under the same conditions show the same results.

Now we go back to our Table and compare SD of our experiments which actually say about our precission (there where measurments are precise the SD is small) and we comment. For example, the experiment A6 is the most precisie and the worts one is experiment A5 etc (of course you should put more than one sentence:)

Part e) As usually the last part is about simply calculations:

c1 = 50microM =

c2 = 5nM

dilution factor = c1 / c2 = 50 000nM / 5nM =

c3 = c1 diluted 500 times = c1 / 500 = 50microM / 500 =

c4 = 5.84% = it means that in 100g of the solution there is

n (number of moles) = mass of the substance / molecular mass

nNaCl = 5.84g / 58.4 g/mol =

Molar Concentration = number of moles / volume

MC of NaCl = 0.1mol / 1L = 0.1mol/L =

We have done it:) CyA Soon:)

Remember that if you have any questions leave the comment on any post or mail me at m.kliszczak1@nuigalway.ie!!

Maciek GGS TEAM

Precision - which might be called reproducubility or repeatabillity as well is the degree to which repeated measurments under the same conditions show the same results.

Now we go back to our Table and compare SD of our experiments which actually say about our precission (there where measurments are precise the SD is small) and we comment. For example, the experiment A6 is the most precisie and the worts one is experiment A5 etc (of course you should put more than one sentence:)

Part e) As usually the last part is about simply calculations:

c1 = 50microM =

**0.050mM**=**50 000nM**c2 = 5nM

dilution factor = c1 / c2 = 50 000nM / 5nM =

**10 000**[no unitis:)]c3 = c1 diluted 500 times = c1 / 500 = 50microM / 500 =

**0.1microM**=**100nM**=**0.0001mM**c4 = 5.84% = it means that in 100g of the solution there is

**5.84g**of the NaCl (100g x 5.84% =**5.84g**)n (number of moles) = mass of the substance / molecular mass

nNaCl = 5.84g / 58.4 g/mol =

**0.1mol**Molar Concentration = number of moles / volume

MC of NaCl = 0.1mol / 1L = 0.1mol/L =

**0.1M**We have done it:) CyA Soon:)

Remember that if you have any questions leave the comment on any post or mail me at m.kliszczak1@nuigalway.ie!!

Maciek GGS TEAM

Excellent blog about calculating molar mass.

ReplyDeleteA very useful guideline to learn how to get success in the school examinations and how to excel in the life.

For any chemical compound that's not an element, we need to find the molar mass from the chemical formula. To do this, we need to remember a few rules:

1. Molar masses of chemical compounds are equal to the sums of the molar masses of all the atoms in one molecule of that compound. If we have a chemical compound like NaCl, the molar mass will be equal to the molar mass of one atom of sodium plus the molar mass of one atom of chlorine. If we write this as a calculation, it looks like this:

(1 atom x 23 grams/mole Na) + (1 atom x 35.5 grams/mole Cl) = 58.5 grams/mole NaCl