You have 50.55uM of MgCl2 which is the same as 0.00295g. You have 12 wells. How

Question

You have 50.55uM of MgCl2 which is the same as 0.00295g. You have 12 wells. How much water is necessary to dilute the sample so that: Well 1 has 0% MgCl2 Well 2 has 1% MgCl2 Well 3 has 10% MgCl2 etc for the remaining wells at 20%,30%,40%,50%,60%,70%,80%,90%,100% MgCl2 respectively. Show your work. You have 50.55uM of MgCl2 which is the same as 0.00295g. You have 12 wells. How much water is necessary to dilute the sample so that: Well 1 has 0% MgCl2 Well 2 has 1% MgCl2 Well 3 has 10% MgCl2 etc for the remaining wells at 20%,30%,40%,50%,60%,70%,80%,90%,100% MgCl2 respectively. Show your work. Well 1 has 0% MgCl2 Well 2 has 1% MgCl2 Well 3 has 10% MgCl2 etc for the remaining wells at 20%,30%,40%,50%,60%,70%,80%,90%,100% MgCl2 respectively. Show your work.

Explanation / Answer

Dilution experiments and preparation of solutions,

we have 0.00295 g (50.55 uM) MgCl2 solution

Well 1 : 0% MgCl2

No MgCl2 solution added to the solution prepared. Only water is present.

Well 2 : 1% MgCl2

1 g in 100 ml makes 1% (w/v) solution

So, 0.00295 g would need = 0.00295 x 100 = 0.295 g (or 0.295 ml) water for dilution

Well 3 : 10% MgCl2

10 g in 100 ml makes 10% (w/v) solution

So, 0.00295 g would need = 0.00295 x 100/10 = 0.0295 g (or 0.0295 ml) water for dilution

Well 4 : 20% MgCl2

20 g in 100 ml makes 10% (w/v) solution

So, 0.00295 g would need = 0.00295 x 100/20 = 0.01475 g (or 0.01475 ml) water for dilution

Well 5 : 30% MgCl2

30 g in 100 ml makes 10% (w/v) solution

So, 0.00295 g would need = 0.00295 x 100/30 = 0.00983 g (or 0.00983 ml) water for dilution

Well 12 : 100% MgCl2

Pure MgCl2 solution. No water was added.

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