± Concentration Molarity (M) is defined as the number of moles of solute divided
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± Concentration Molarity (M) is defined as the number of moles of solute divided by the solution volume expressed in liters molarity oumof solution (L) For example, 1 HCl contains 1 mol of HCl dissolved in 1 L of the water. When a concentrated solution is diluted number of moles of solute stays constant, only the volume of the solution is changed. A dilution indicates an increase in solution volume and, therefore, the concentration of the solution must decrease. If you add more water to the considered above, so that now the volume is 2 L. the molarity of the solution is now 1 mol in a 2 L solution, that is, (1/2) M or 0.5 M. HCl solution the number of moles remains the same but the volume is doubled. Hence The number of moles of solute before and after dilution can be calculated by multiplying molarity times volume. We can set up the following equations moles of solute molarity × volume where M, is the initial molarty (of the concentrated solution), Vi isthe initial volume, M is the final molarity (of the diluted solution), and Vr is the final volume. , Vi is the initial volume, Me is the final molarity (of the diluted In the HCl solution example the initial molarity is 1 M, the initial volume is 1 L, the finai volume is 2 L, and the molarity is 0.5 M. Thus the number of moles present in these solutions is Part A The following five beakers, each containing a solution of sodium chioride (NaCl, also known as table salt), were found on a lab shelf Beaker Contents 1 200. mL of 1.50 M NaCI solution 2 100. mL of 3.00 M NaCl solution 3 150 mL of solution containing 23.0 g of NaCI 4 100. mL of solution containing 23.0 g of NaCI 5 300. mL of solution containing 0.450 mol NaCI Arrange the solutions in order of decreasing concentration. Rank from most concentrated to least concentrated. To rank items as equivalent, overlap them You did not open hints for this part ANSWERExplanation / Answer
Part-A:
NaCl concentration of beaker-1 = 1.50 M
NaCl concentration of beaker-2 = 3.00 M
For beaker-3: Volume, V = 150 mL = 150 mL x (1L / 1000 mL) = 0.150 L
moles of NaCl = mass / molar mass = 23.0 g / 58.44 g/mol = 0.3936 mol
Hence concentration of NaCl in beaker-3 = moles of NaCl / V(L) = 0.3936 mol / 0.150 L = 2.624 M
For beaker-4: Volume, V = 100 mL = 100 mL x (1L / 1000 mL) = 0.100 L
moles of NaCl = mass / molar mass = 23.0 g / 58.44 g/mol = 0.3936 mol
Hence concentration of NaCl in beaker-4 = moles of NaCl / V(L) = 0.3936 mol / 0.100 L = 3.936 M
For beaker-5: Volume, V = 300 mL = 300 mL x (1L / 1000 mL) = 0.300 L
moles of NaCl = 0.450 mol
Hence concentration of NaCl in beaker-5 = moles of NaCl / V(L) = 0.450 mol / 0.300 L = 1.50 M
Hence the rder of concentration is
Beaker-4 (3.936 M) > Beaker-2 (3.00 M) > Beaker-3 (2.624 M) > Beaker-1 (1.50 M) = Beaker-5 (1.50 M)
(Most concentrated) ---------------------------------------------------------------------- (Least concentrated)
Note that beaker-1 and beaker -5 have same concentration and they need to be overlapped.
Part-B:
moles of glucose taken = mass / molar mass = 18.0 g / 180.0 g/mol = 0.10 mol
Volume of the sample solution, V = 100 mL = 0.100 L
Hence concentration of the sample solution, M1 = moles of glucose / V(L) = 0.10 mol / 0.100 L = 1.0 M
Dilution:
Volume of sample solution taken, V1 = 50 mL = 0.050 L
concentration of sample solution, M1 = 1.0 M
Volume of solution after dilution, V2 = 0.500 L
Let the concentration of the solution after dilution be 'M2'
Applying law of dilution
M1V1 = M2V2
=> M2 = M1V1/V2 = (1.0 Mx 0.050 L) / 0.50L = 0.1 M
now moles of glucose in 0.500L of the diluted solution = M2xV2 = 0.1 M x 0.5L = 0.05 mol
Hence moles of glucose in 100 mL (0.100L) of diluted solution = 0.05 mol x (0.100L / 0.500L) = 0.01 mol
Hence mass of glucose = 0.01 mol x 180 g/mol = 1.8 g (answer)
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