Molarity (or molar concentration) is one of several possible ways to express the

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Molarity (or molar concentration) is one of several possible ways to express the solution concentration. Since molarity includes the number of moles of solute (in the numerator of the unit), it is very useful in stoichiometry calculations for reactions taking place in solution. The unit of molarity, M, can also be written as mol/L. Molarity (M) is defined as the number of moles of solute divided by the solution volume expressed in liters moles of solute volume of solution (L) molarity- Part B For example, 1 M HCl contains 1 mol of HCI dissolved in 1 L of the water. When a concentrated solution is diluted, the 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 HCI solution considered above, so that now the volume is 2 L, the number of moles remains the same but the volume is doubled. Hence the molarity of the solution is now 1 mol in a 2 L solution, that is, (1/2) M or 0.5 M A student placed 11.5 g of glucose C126) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. ImL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 40.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100. mL. of the final solution? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) 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: 0.4590.459 gg moles of solute molarity x volume = where Mi is the initial molarity (of the concentrated solution), Vi is the initial volume, Mr is the final molarity (of the diluted solution), and V is the final volume Submit In the HCl solution example the initial molarity is 1 M the initial volume is 1 L, the final volume is 2 L, and the molarity is 0.5 M. Thus the number of moles present in these solutions is X incorrect. Try A gain; 5 attempts remaining Provide Feedback Next >

Explanation / Answer

Determine the molar mass of glucose, C6H12O6. The atomic masses are

C: 12.011 u

H: 1.008 u

O: 15.999 u

The gram molar mass of C6H12O6 = (6*12.011 + 12*1.008 + 6*15.999) g/mol = 180.156 g/mol.

Moles of glucose corresponding to 11.5 g = (mass of glucose)/(molar mass of glucose) = (11.5 g)/(180.156 g/mol) = 0.06383 mole.

The student dissolved 11.5 g glucose in a final volume of 100 mL. The volume of the solution in liters = (100 mL)*(1 L/1000 mL) = 0.100 L.

Molarity of the prepared glucose solution = (moles of glucose)/(volume of the solution in L) = (0.06383 mole)/(0.100 L) = 0.6383 mol/L.

40.0 mL of the prepared glucose solution was taken and diluted to a final volume of 0.500 L. Use the dilution equation to determine the molarity of glucose in the new solution (0.500 L). The dilution equation is

C1*V1 = C2*V2

where C1 = concentration of stock (original solution) = 0.6383 mol/L; V1 = volume of stock solution taken = 40.0 mL; V2 = volume of the new solution prepared = 0.500 L and C2 = concentration of the new solution. Therefore,

(40.0 mL)*(0.6383 mol/L) = (0.500 L)*C2

====> C2 = (40.0 mL)*(1 L/1000 mL)*(0.6383 mol/L)/(0.500 L) = 0.051064 mol/L.

Moles of glucose in 100 mL of the new solution = (volume of the solution in L)*(molarity of the solution) = (100 mL)*(1 L/1000 mL)*(0.051064 mol/L) = 0.0051064 mole.

Mass of glucose corresponding to 0.0051064 mole glucose = (moles of glucose)*(molar mass of glucose) = (0.0051064 mole)*(180.156 g/mol) = 0.91995 g ? 0.92 g (correct to three sig. figs, ans).

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