The Mole Concept Exam1 and Problem Solutions

The Mole Concept Exam1 and  Problem Solutions

 

1. If atomic mass of Mg atom is 24 g, find mass of 1 Mg atom.

Solution:

We can solve this problem in to ways;

1st way:

6,02x1023 amu is 1 g

24  amu           is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=4x10-23 g

2nd way;

1 mol Mg (6,02x1023 Mg atoms) is 24 g

6,02x1023 Mg atoms   24 g

1              Mg atom     ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=4x10-23 g

 

2. Find mass of 1 molecule C2H6. (C=12, H=1)

Solution:

C2H6=2.12+6.1=30

6,02x1023 C2H6 molecule is 30 g

1              C2H6 molecule is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

1 C2H6 molecule is =?=5.10-23 g

 

3. Find mole of 6,9 g Na. (Na=23)

Solution:

23 is the atomic mass of Na, in other words 1 mole Na is 23 g.

23 g Na is 1 mol

6,9 g Na is ? mol

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,3 mol

 

4. Find mass of 0,2 mol P4 . (P=31)

Solution:

Molecule mass of P4 =4.31=124 g

1 mol P4 is 124 g

0,2 mol P4 is ? mol

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?= 24,8 g

 

5. Find mole of 4,48 liters O2 under normal conditions.

Solution:

Under normal conditions, 1 mol gas is 22,4 liters. We use following formula to find moles of gas under normal conditions;

n=V/22,4

n=4,48/22,4=0,2 mol

 

6. Find mass of Fe in the compound including 4,8x1023 O atoms ;Fe3O4 .

Solution:

We first find mole of O in the compound;

nO=(4,48x1023)/(6,02x1023)=0,8mol

Now we find mole of compound that contains 0,8mol O;

nFe3O4=(moles of O)/(moles of O in compound)=0,8/4=0,2mol

Mole of Fe in compound is;

There are 3 mol Fe in 1 mol Compound

there are ? mol Fe in 0,2mol compound

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=0,6 mol Fe in compound

mass of 0,6 mol Fe is;

1 mol Fe is 56 g

0,6 mol Fe is ? g

¯¯¯¯¯¯¯¯¯¯¯¯¯

?=33,6g There are 33,6 g Fe in compound

 

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