**Question 1 (00000001A0601402, sports No. 70): **113g **Feedback**You have to refer come a regular table and look up theatomic load of chromium. Girlfriend will uncover it to it is in 52.00 (rounded to foursignificant figures). This way that:

*The ***molarmass*** the chromium is 52.00 g/mol.You are watching: How many atoms are in 1 mole of silver*

**One mole**

*the chromium atoms has a massive of 52.00 g*

Therefore: n mole of chromiumatoms will have a fixed of (n times 52.00) g.OR: 2.18 mole of chromium atom will have a mass of (2.18 times 52.00) g,which is same to 113 g. Here"s just how it"s done using dimensional analysis:

2.18 mol chromium atoms | 52.00 g chromium | |||

x | | = | 113 g chromium | |

1 mol chromium atoms |

NOTES:1. "mol" is the main abbreviation because that "mole" (or"moles"). 2. The number of far-ranging figures in our molar mass should, if possible,exceed the in the measure mass. Due to the fact that there are 3 far-reaching figures in 2.18g; the is enough to round off our molar mass to four far-ranging figures(52.00 g/mol).

**Question 2 (00000001A0601403, variation No. 53): **0.0988mol sulfur atoms **Feedback**You need to refer come a routine table and also look up the atomic weight of sulfur.You will uncover it to it is in 32.07 (rounded to four far-reaching figures). This meansthat:

*The ***molarmass*** of sulfur is 32.07 g/mol.***One mole*** of sulfur atoms has a massive of 32.07 g*

Therefore: n moles of sulfuratoms will have actually a massive of (n times 32.07) g.The problem says: 3.17 g equates to (n times 32.07) g; therefore we fix for n bydividing 3.17 by the molar mass (32.07). Here"s exactly how it"s done utilizing dimensional analysis:

3.17 g sulfur atoms | 1 mol sulfur atoms | |||

x | | = | 0.0988 mol sulfur atoms | |

32.07 g sulfur atoms |

**Question 3 (00000001A0601404, sport No. 57): **3.84x1022silver atom **Feedback**You must refer to a regular table and also look increase the atomic load of silver.You will uncover it to be 107.9 (rounded come four significant figures). This meansthat:

*The ***molarmass*** of silver is 107.9 g/mol.See more: What Is The Study Of Matter And The Changes It Undergoes ? What Is The Branch Of Science That Deals With The*

**One mole**

*of silver- atoms has actually a mass of 107.9 g*

Therefore: n mole of silveratoms will have actually a massive of (n times 107.9) g.The problem says: 6.89 g amounts to (n times 107.9) g; so we resolve for n bydividing 6.89 by the molar massive (107.9). This (n) offers us the mole count. Toget the individual atom count, we multiply n through Avogadro"s number. Here"s exactly how it"s done utilizing dimensional analysis. First, us calculate the variety of moles:

6.89 g silver atoms | 1 mol silver atoms | |||

x | | = | 0.06386 mol silver- atoms | |

107.9 g silver atoms |

THEN, we transform mole count to separation, personal, instance atom count:

0.06386 mol silver- atoms | 6.02x1023 silver- atoms | |||

x | | = | 3.84x1022 silver atoms | |

1 mol silver- atoms |

Here"s a an ext direct approach, since 1 mole of atom is 6.02x1023atoms:

6.89 g silver atoms | 6.02x1023 silver atoms | |||

x | | = | 3.84x1022 silver atoms | |

107.9 g silver- atoms |

**Question 4 (00000001A0601405, sports No. 28): **1.55x102g silver atoms **Feedback**You must refer to a periodic table and look up the atomic weight of silver.You will uncover it to be 107.9 (rounded to four significant figures). This meansthat:

*The ***molarmass*** of silver is 107.9 g/mol.***One mole*** of silver atoms has actually a massive of 107.9 g*

Therefore: n moles of silveratoms will have a mass of (n time 107.9) g. We have the right to calculate (n) through simplydividing the separation, personal, instance atom counting by Avogadro"s number. Climate multiply (n) bythe molar fixed to obtain the answer. Here"s just how it"s done utilizing dimensional analysis. First, we transform individual atom counting to mole count: