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Paired Exercise 14.18 Calculate the molarity of the following solutions: (a) 0.67 mol of solute in 115 mL of solution M the tolerance is +/-2% (b) 5.08 mol of CaCl, in 8.73 L of solution M CaCl2 the tolerance is +/-2% (c) 275 g of C6H12O6 in 775 mL of solution M C6H1206

Paired Exercise 14.18 Calculate the molarity of the following solutions: (a) 0.67 mol of solute in 115 mL of solution M the tolerance is +/-2% (b) 5.08 mol of CaCl, in 8.73 L of solution M CaCl2 the tolerance is +/-2% (c) 275 g of C6H12O6 in 775 mL of solution M C6H1206

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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### Paired Exercise 14.18 **Calculate the molarity of the following solutions:** --- #### (a) 0.67 mol of solute in 115 mL of solution \[ \text{Molarity (M)} = \] *the tolerance is +/- 2%* --- #### (b) 5.08 mol of CaCl₂ in 8.73 L of solution \[ \text{Molarity (M)} \, \text{CaCl₂} = \] *the tolerance is +/- 2%* --- #### (c) 275 g of C₆H₁₂O₆ in 775 mL of solution \[ \text{Molarity (M)} \, \text{C₆H₁₂O₆} = \] *the tolerance is +/- 2%* --- #### (d) 144. g of MgSO₄ · 7 H₂O in 4.71 L of solution \[ \text{Molarity (M)} \, \text{MgSO₄} = \] *the tolerance is +/- 2%* --- In this exercise, you are tasked with calculating the molarity, which is the number of moles of solute per liter of solution. Make sure to convert all volumes to liters and mass to moles where necessary. Then use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] Ensure your final answer falls within the provided tolerance.
Transcribed Image Text:### Paired Exercise 14.18 **Calculate the molarity of the following solutions:** --- #### (a) 0.67 mol of solute in 115 mL of solution \[ \text{Molarity (M)} = \] *the tolerance is +/- 2%* --- #### (b) 5.08 mol of CaCl₂ in 8.73 L of solution \[ \text{Molarity (M)} \, \text{CaCl₂} = \] *the tolerance is +/- 2%* --- #### (c) 275 g of C₆H₁₂O₆ in 775 mL of solution \[ \text{Molarity (M)} \, \text{C₆H₁₂O₆} = \] *the tolerance is +/- 2%* --- #### (d) 144. g of MgSO₄ · 7 H₂O in 4.71 L of solution \[ \text{Molarity (M)} \, \text{MgSO₄} = \] *the tolerance is +/- 2%* --- In this exercise, you are tasked with calculating the molarity, which is the number of moles of solute per liter of solution. Make sure to convert all volumes to liters and mass to moles where necessary. Then use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \] Ensure your final answer falls within the provided tolerance.
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