Question

If water has a freezing point depression constant of 1.86 °C/m, what temperature would a 15.0 % solution of Magnessium Nitrate freeze?

Answer 1

Following is the - complete Answer -&- Explanation, for the given Question, in.....typed format...

Answer:

Freezing point of 15% w/v aqueous solution of Mg(NO₃)₂ = - 5.64 ^oC

Explanation:

Following is the complete Explanation, for the above Answer...

• Step - 1:

We know the following formula, for determining the value of freezing point depression, of the aqueous solution, of a compound:

T_f = i  x K_f  x m   ------------------------------------- Equation - 1

Where:

1. T_f= freezing point depression value [=] ^oC
2. i =   van't Hoff factor of the aqueous solution of a solute [=] unitless
3.   K_f = freezing point depression constant [=] ^oC / m
4. m =   molality of the solute [=] mol of solute/ kg of dissolving solvent

• ​​​​​​​Step - 2:

​​​​​​​For the given solution of magnesium nitrate ( Mg(NO₃)₂) , we know the following:

1. i = van't Hoff factor =  3.0
2.    K_f = Freezing point depression constant for Water as solvent = 1.86 ^oC/m

​​​​​​​We will have to find the molality, of the given: 15% w/v solution, of Mg(NO₃)₂  , in order to plug the value into   Equation - 1, ...determined as the following:

• Step - 3:

​​​​​​​Let's assume, that we have an 15.0 % w/v solution, of Mg(NO₃)₂ :

i.e. 15.0 g ( grams ) of  Mg(NO₃)₂ is dissolved in 100.0 mL  of Water .

OR,

150.0 g ( grams ) of Mg(NO₃)₂ is dissolved in 1000.0 mL  of Water

We know the density of water ( at room temperature ) is approximately =  1.0 g/mL ( i.e. grmas per mL )

Therefore: mass of 1000.0 mL water will be = ( 1000.0 mL ) x ( 1.0 g/mL ) = 1000.0 g OR = 1.0 kg solvent...

• Step - 4:

​​​​​​​Using the above information, we can derive/ calculate the following:

1. Number of moles of Mg(NO₃)₂ = ( 150.0 g ) / ( 148.3 g/mol ) =  1.011 mol ( moles )
2. Mass of solute ( water ) =  1.0 kg  

​​​​​​​Therefore, molality of the given solution = ( 1.011 mol ) / ( 1.0 kg ) =  1.011 m

• Step - 5:

​​​​​​​Therefore, plugging in values in Equation - 1, we will get the following:

   T_f = i  x K_f  x m

  T_f = ( 3.0 )  x ( 1.86 ^oC/m ) x ( 1.011 m ) =  5.64 ^oC

• Step - 6:

​​​​​​​Therefore, the following will be the freezing point, of the given aqueous solution of Mg(NO₃)₂ ...

Freezing point = ( 0.0  ^oC ) - ( 5.64 ^oC ) =  - 5.64 ^oC ....

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