# Dilution calculation for 10 ml of 2 M stock solution. Molecular weight for the component is 48.6 g / mol.

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From the information you give it sounds like you want to know the grams of the substance contained in the given sample. If you have a 2 M solution, that means that it contains 2 moles of chemical per liter of solution. If we have 10 mL of this solution then we convert this to liters (it is 0.01 liters) and multiply the two numbers to find the number of moles in 10 mL.

2 mol/L * (0.01 L) = 0.02 moles of solute

Since we know that the solute has a molecular weight of 48.6 g/mol, we can again multiply to find the number of grams in the 10 mL sample.

0.02 moles * (48.6 g/mol) = 0.972 grams solute

So we have found that there are 0.972 grams of the solute contained in the 10 mL sample of the solution.

To what concentration you want to dilute the solution?

if you dilute 10mL 2M solution to 100 mL you would get a 0.2M solution.

use formula

c1v1=c2v2

for above problem

stock=dilution

c1v1=c2v2

c1=(c2*v2)/v1

here c1=?,c2=2M,v1=100mL & v2=10mL

c1=(2*10)/100

c1=0.2M