Hey everyone
Just having a bit of trouble with this dilution question, any help would be appreciated, thnx in advance
' The concentration of a solution of ammonia (NH3) is 1.5%(m/v). What is the molar concentration of a solution produced by diluting 25.0 mL of this solution with 250 mL of water?
First we need to find the moles present in the 25.0mL solution of 1.5% m/v NH
3. We need to remember that 1.5% m/v = 1.5g of NH
3 in every 100mL
We multiply our target (25.0 mL) by our current ratio (1.5 g NH
3 / 100 mL ) to get 0.375 g of NH
3Using one of our favourite formulas (n=m/M), we get 0.02202 moles of NH
3 in our 25.0mL solution
Remember that as we're diluting by adding water, the number of moles of solute stays the same. So we have 25.0 mL currently, and we add 250 mL to that, so we have 275 mL of solution total which contains 0.02202 moles of NH
3Using our formula for molarity (moles of solute/litres of solution = 0.02202 moles / 0.275 L), we get 0.080 mol L
-1Two sig figs, so the answer is 0.08 mol L
-1