Question

7.3g HCl in 100ml solution. Write down the concentration of the solution in g/L and mol/L.

Answer 1

"Given:  

7.3g HCl in 100ml solution. (Molar mass of HCl is 36.46g/mol)

To find:  

The concentration of the solution in g/L and mol/L.

Formula to be used:

$Grams\quad per\quad liter\quad =\quad \frac { Mass\quad of\quad solute }{ Volume\quad of\quad solution }$

$=\quad \frac { 7.3\quad g }{ 1000\quad ml } \quad =\quad \frac { 7.3\quad g }{ 1\quad \left( as\quad 1000\quad ml\quad is\quad 1\quad liter \right)  }$

$=\quad 7.3\quad { g }/{ L }$

$Moles\quad per\quad liter\quad =\quad \frac { Moles\quad of\quad solute }{ Volume\quad of\quad solution }$

Solution:

$Grams\quad per\quad liter\quad =\quad \frac { Mass\quad of\quad solute }{ Volume\quad of\quad solution }$

$=\quad \frac { 7.3\quad g }{ 1000\quad ml } \quad =\quad \frac { 7.3\quad g }{ 1\quad \left( as\quad 1000\quad ml\quad is\quad 1\quad liter \right)  }$

$=\quad 7.3\quad { g }/{ L }$

In order to find the composition of 1 mole of HCl in the solution the following step is done:

$7.3\quad g\quad \times \quad \frac { 1\quad mol\quad of\quad HCl }{ 36.46\quad \cfrac { g }{ mol } \quad \left( Molar\quad mass\quad of\quad HCl \right)  }$

1 Mol of HCl contains = 0.200 moles HCl

Substituting the obtained mole value in the molarity formula we get:

$Moles\quad per\quad liter\quad =\quad \frac { Moles\quad of\quad solute }{ Volume\quad of\quad solution }$

$=\quad \frac { 0.200 }{ 1000 } \quad =\quad \frac { 0.200 }{ 1\quad \left( 1000\quad ml\quad is\quad 1\quad liter \right)  }$

$=\quad 0.2\quad { mol }/{ L }$

Conclusion:

The concentration of the solution in:

Grams per liter is 7.3 g/L

Moles per liter is 0.2 mol/L"

Answer 2

Explanation:

7.3g HCl in 100ml solution. Write down the concentration of the solution in g/L and mol/L.

7.3g HCl in 100ml solution. (Molar mass of HCl is 36.46g/mol)

To find:

The concentration of the solution in g/L and mol/L.

Formula to be used:

Grams\quad per\quad liter\quad =\quad \frac { Mass\quad of\quad solute }{ Volume\quad of\quad solution }Gramsperliter=

Volumeofsolution

Massofsolute

=\quad \frac { 7.3\quad g }{ 1000\quad ml } \quad =\quad \frac { 7.3\quad g }{ 1\quad \left( as\quad 1000\quad ml\quad is\quad 1\quad liter \right) }=

1000ml

7.3g

=

1(as1000mlis1liter)

7.3g

=\quad 7.3\quad { g }/{ L }=7.3g/L

Moles\quad per\quad liter\quad =\quad \frac { Moles\quad of\quad solute }{ Volume\quad of\quad solution }Molesperliter=

Volumeofsolution

Molesofsolute

Solution:

Grams\quad per\quad liter\quad =\quad \frac { Mass\quad of\quad solute }{ Volume\quad of\quad solution }Gramsperliter=

Volumeofsolution

Massofsolute

=\quad \frac { 7.3\quad g }{ 1000\quad ml } \quad =\quad \frac { 7.3\quad g }{ 1\quad \left( as\quad 1000\quad ml\quad is\quad 1\quad liter \right) }=

1000ml

7.3g

=

1(as1000mlis1liter)

7.3g

=\quad 7.3\quad { g }/{ L }=7.3g/L

In order to find the composition of 1 mole of HCl in the solution the following step is done:

7.3\quad g\quad \times \quad \frac { 1\quad mol\quad of\quad HCl }{ 36.46\quad \cfrac { g }{ mol } \quad \left( Molar\quad mass\quad of\quad HCl \right) }7.3g×

36.46

mol

g

(MolarmassofHCl)

1molofHCl

1 Mol of HCl contains = 0.200 moles HCl

Substituting the obtained mole value in the molarity formula we get:

Moles\quad per\quad liter\quad =\quad \frac { Moles\quad of\quad solute }{ Volume\quad of\quad solution }Molesperliter=

Volumeofsolution

Molesofsolute

=\quad \frac { 0.200 }{ 1000 } \quad =\quad \frac { 0.200 }{ 1\quad \left( 1000\quad ml\quad is\quad 1\quad liter \right) }=

1000

0.200

=

1(1000mlis1liter)

0.200

=\quad 0.2\quad { mol }/{ L }=0.2mol/L

Conclusion:

The concentration of the solution in:

Grams per liter is 7.3 g/L

Moles per liter is 0.2 mol/L"