For each of the following solutions, calculate the molarity of the solution.9.71 g MgNH4PO4 in 300. mL solution mol/L12.6 g NaCH3COO in 500. mL solution mol/L1.51 g CaC2O4 in 400. mL solution mol/L4.96 g (NH4)2SO4 in 300. mL solution mol/L

Introduction Molarity: Molarity is defined as number of moles of the solute in 1 litres of the solution Molarity= Moles(solute)Volume in L(solution) Step:1 1.9.71 g MgNH4PO4 in 300 mL solution 300 mL= 0.300 L MM of MgNH4PO4 = 137.31g/mol Moles of MgNH4PO4 = 9.71g137.31g/mole=0.07 mole Molarity = 0.07 mole0.300L=0.233 M 2.12.6 g NaCH3COO in 500 mL solution 500 mL=0.500L MM of NaCH3COO=82.03g/mole Moles of NaCH3COO= 12.6g82.03g/mol=0.1536 mole Molarity= 0.1536 mole0.500L=0.3072 M3.1.51 g of CaC2O4 in 400 mL solution 400 mL=0.400 L MM of CaC2O4 =128.097g/mol Moles of CaC2O4 = 1.51g128.097g/mol=0.01178 mole Molarity=0.01178 mole0.400 L=0.02945M 4.4.96 g (NH4)2SO4 IN 300mL solution 300mL =0.300 L MM of (NH4)2SO4 =132.14g/mol Moles of (NH4)2SO4 = 4.96g132.14g/mol=0.0375 mole Molarity = 0.0375 mole0.300L=0.125 M

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Chemistry

10th Edition

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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For each of the following solutions, calculate the molarity of the solution.

9.71 g MgNH₄PO₄ in 300. mL solution

mol/L
12.6 g NaCH₃COO in 500. mL solution

mol/L
1.51 g CaC₂O₄ in 400. mL solution

mol/L
4.96 g (NH₄)₂SO₄ in 300. mL solution

mol/L

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