Continuing from where I left off in part one, we are faced with yet another equation and type of question.

6. **You dissolve 17 g of silver nitrate in 100 ml. What is the molarity? Silver nitrate mwt = 169.87**

*Molarity is defined as the concentration of a solution in terms of moles of substance per litre. *

Now molarity has it’s own equation, which is essentially two equations. Initially, I worked out the two separately and ended up with the wrong answer. I will tell you what I did wrong and highlight the importance of knowing the difference between molarity and concentration.

Firstly, molarity = mass/molar mass/vol. That’s kind of hard to write here but basically, you divide mass by the molar mass (which is the molecular weight) and then divide the answer of the by the volume in **litres. **The mistake I made when I first did this question was to leave the volume in ml. For molarity, volume must always be in **litres .**

Okay, so we have 17g of silver nitrate in 100ml. We are told that the molecular weight (mwt) of silver nitrate is 169.87.

First, we do mass/molar mass = 17g/169.87. If you’re good at mental maths, you’ll see that the mwt basically rounds up to 170 and quickly not that the answer is 0.1 mol. If you’re not good at mental maths, your calculator will help you out!

After working out the moles, convert the volume in the question to litres. In this case we have 100ml. As 1000ml makes 1L, we can work out that 100ml = 100/1000 = 0.1L

Taking our answer from earlier (0.1 moles), we then do that divided by our converted volume (0.1ml). So 0.1/0.1 = 1M. The unit for molarity is **M**. So the molarity is 1.

7. You need to make 2 L of a 5 M solution of NaCl. Mwt of NaCl = 58.44. What mass do you need?

Using the above equation for molarity, we know that:

**molarity = mass/molar mass/vol. **

To work out the mass in this instance we need to rearrange the equation by first multiplying both sides by volume and then the molar mass, leaving us with:

**mass = molar mass x vol x molarity**

We then go back to the question and see what information we have. We know that 2L = volume, 5M = molarity and molar mass = 58.44, so the mass required must equal 58.44 x 2 x 5 = 584.4g

8. You take 10 ml of 5 M NaCl and make up to a 100 ml solution. What is the final concentration?

For this question, we use yet another equation:

**C1V1 = C2V2**,

*where C1 = initial concentration, V1 = initial concentration, C2 = final concentration and V2 = final volume*

So, from what we’ve been told we can see that C1 = 5, V1 = 10 and V2 = 100. We put that into the equation and get:

5 x 10 = 100 x C2

50 = 100 x C2. To work out C2 we will have to divide both sides by 100

50/100 = C2

0.5 = C2, therefore the final concentration in this instance is 0.5M

Very exciting!!!

9. **You need to make 500 ml of 70% ethanol solution. How much pure ethanol do you need?**

A 70% solution means that 70% of the reagent makes up the final volume. So in out 500ml, 70% of that is ethanol and the remaining 30% will made up with distilled water (solution usually = reagent + distilled water).

70% of 500ml = 350ml = amount of pure ethanol needed. The remaining 150ml of the solution will be made up with distilled water.

** If ethanol stock was actually 90% ethanol then how much would you need?**

This just means that the ethanol stock isn’t 100%, so in the above question when we worked out 70% of 500ml, we techincally did 70/100 x 500. In this case we need to do 70/90 x 500 = 389ml ethanol needed

I struggled with this one but I think it’s just a case of knowing what % reagent really means.

10. Make 500 ml of a 20% w/v SDS solution. How much SDS do you need?

% solutions w/v = weight by volume solutions . They are used for a solution of a solid.

We have to assume that 1 ml of the solution has mass 1 g.

So mass needed = % of the mass of the solution (100ml = 100g)

In this question, we basically assume that 500ml = 500g.

20% of 500g = 100g. So 100g of SDS needed.

(Bonus fact: SDS is **Sodium dodecyl sulfate **a surfactant used in cleaning products and also different biological application methods, like separating proteins (e.g. DNA).)