Question

If 0.015 mg Pb2+ were added to enough water to make exactly 1 L of solution, the solution would be ____ M in Pb2+

7.2x10-3 |

1.5x10-5 |

7.2x10-8 |

1.5x10-8 |

7.2x10-5 |

7.2x103 |

2. The MCL for Pb2+ is 0.015 ppm this is equivalent to ______

15 mg/L |

1.5 mg/L |

0.15 mg/L |

0.015 mg/L |

0.0015 mg/L\]\ |

Answer #1

**Answer – 1)** We are given, mass of
Pb^{2+} = 0.015 mg, volume = 1.0 L

Molarity of solution = ?

We need to calculate the moles of Pb^{2+}

We know,

1 mg = 0.001 g

So,0.015 mg = ?

= 1.5*10^{-5} g

moles of Pb^{2+} = 1.5*10^{-5} g / 207.2
g.mol^{-1}

= 7.24*10^{-8} moles

We know molarity is

Molarity = moles / L

= 7.24*10^{-8} moles / 1.0 L

= 7.24*10^{-8} M

So, the solution would be **7.2*10 ^{-8} M**
in Pb

**2)** [Pb^{2+}] = 0.015 ppm

We know

1 ppm = 1 mg/ L

So, 0.015 ppm = 0.015 mg/L

So, the MCL for Pb2+ is 0.015 ppm this is equivalent to
**0.015 mg/L**

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