#5 teaching martians maths for bio-scientists (part one)

According to my tutor, I need to approach questions in a way that assumes the reader has no idea what I’m talking about. Or pretend that I explaining something to a Martian. Yeah. Kinda weird, but I’m willing to give it a try because blogging is keeping me awake.

So today, I will be teaching you and the Martians some basic calculations.

1. We dissolve 20g of glucose in 100ml of distilled water. What is the final concentration of glucose?

Well, first of all we know how much glucose we have per 100ml – 20g.

The formula for concentration = mass/volume

So to work out the concentration we do 20g/100ml, which is 0.2g/ml. I’m fairly certain that the answer can be left like that, but if in doubt, the 20g can be converted into mg. Or wait, actually, I think converting ml to L would be a lot better. BUT ANYWAY, NEXT QUESTION.

2. You have weighed out 25 mg of an enzyme and need to make a 10 mg ml-1 What volume of solution can you make?

In this case, the equation conc = mass/vol will need to be rearranged. The question gives us the mass and the concentration, so it will be changed so that volume is what we calculate. To do that, you’d have to multiply both sides of the equation by the volume , making the rearranged concentration conc x vol = mass. We then have to rearrange it again by dividing mass by concentration to have volume on its own, making the equation

vol = mass/conc

Going back to the question, we were given a mass of 25mg and a concentration of 10 mg ml-1.

volume = 25mg/10mg ml-1 = 2.5 ml. The mgs cancel each other out and we’re left with the ml part of the concentration.

This is FUN, right? Yes…

3.  You need to make 100 ml of a 5 mg ml-1 solution of sucrose. How much sucrose do you need?

In this question, we are calculated the mass of sucrose. We have been provided with the volume and concentration, so we have to use the version of the equation wherein mass is what we are trying to calculate. From above, we can see that the initial equation can be rearranged into: conc x vol = mass

Therefore in this case, mass = 5 mg ml-1 x 100 ml = 500mg. Like above, the ‘ml’ part of the equation will cancel out, leaving us with mg. 5 x 100 = 500, giving us an answer of 500mg of sucrose.

4. You need 10 moles of potassium acetate. The mwt is 98.14. What mass do you need?

For the question, the equation we use changes. Moles are the SI unit of amount of substance of a chemical system that contains as many elementary unit as there are atoms in 0.012 kg of carbon-12. Symbol: mol. I’ll be honest with you, I don’t really get that either but you don’t need to in order to find the answer. We turn to another equation

mass = Mr (molecular weight) x n (moles)

That’s it! In this question, we’ve been giving the molecular weight and moles so the mass simply equals 10 x 98.14 which = 981.4g (assume the the weight is in grams, unless the question states otherwise).

5. You weigh out 90 g of glucose. The mwt is 180.16. How many moles do you have?

Here, using the above equation, we can calculate the amount of moles very easily. The equation is rearranged like before into n = mass/Mr. This makes the answer 90/180.16 = 0.498 moles, which we can round up to 0.5 moles.

That’s it for now, I’ll be back with part two soon!

Note: This is purely for my own education purposes, but hey, if you want to learn something new, by my guest!


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