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Molarity | 23 mins | 0 completed | Learn Summary |

Normality & Equivalent Weight | 24 mins | 0 completed | Learn Summary |

Solution Stoichiometry | 22 mins | 0 completed | Learn |

Solubility Rules | 7 mins | 0 completed | Learn Summary |

Net Ionic Equations | 21 mins | 0 completed | Learn Summary |

Electrolytes | 19 mins | 0 completed | Learn Summary |

Redox Reaction | 32 mins | 0 completed | Learn Summary |

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Net Ionic Equation |

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Oxidation Number |

Solution: Which of the following would be used to make a 0.40 M NaCl solution (NaCl mass and water volume, respectively)A. 29.2 g, 500 mL B. 58.4 g, 4.00 L C. 11.7 g, 400 mL D. 58.4 g, 1.00 L E. 5.84 g, 250 mL

Which of the following would be used to make a 0.40 M NaCl solution (NaCl mass and water volume, respectively)

A. 29.2 g, 500 mL

B. 58.4 g, 4.00 L

C. 11.7 g, 400 mL

D. 58.4 g, 1.00 L

E. 5.84 g, 250 mL

Determine which among the choices will produce 0.40 M NaCl using the formula for molarity

Recall that **molarity** is

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

Calculate each molarity produced by each choice where the molar mass of NaCl is:

NaCl** ****1** Na × 22.99 g/mol Na = 22.99 g/mol** 1**

**A. 29.2 g, 500 mL** (1 mL = 1x10^{-3} L)

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