Question

calculate the molality of a solution formed by mixing 500 gram solution of 25 percent.w/w NaOH and 500 gr solution of 15 percent w/w NaOH​

Answer 1

Solution :-

As per the given question ,

Part I :

• Mass of the solution (M 1 ) = 500 g

• %W/W = 25 %

we know that ,

➽ %W/W = mass of NaOH / mass of the solution

➽ 25 = m x 100 / 500

➽ m = 25 x 5

➽ m = 125 g

• Mass of NaOH (m)= 125 g

• Molar mass of NaOH(W) = 40 g

we know that,

➽ n = m / W

➽ n = 125 / 40

➽ n = 3.125

Moles of NaOH  in 500 g solution 1 = 3. 125

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Part II :

• Mass of the solution (M 2 ) = 500 g

• %W/W = 15 %

we know that ,

➽ %W/W = mass of NaOH / mass of the solution

➽ 25 = m' x 100 / 500

➽ m '= 15 x 5

➽ m '= 75 g

• Mass of NaOH (m)= 75 g

• Molar mass of NaOH(W) = 40 g

we know that,

➽ n ' = m '/ W

➽ n '= 75/ 40

➽ n '= 1.875

Moles of NaOH  in 500 g solution 2= 1.875

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Part III :-

we know that ,

➽ m = n + n' / M 1 + M 2

➽ m = 3.125 + 1.875 / 500 + 500

➽ m = 5 / 1000

➽ m = 5 / 1

➽ m = 5 mol / kg

The molality of the solution is 5 mol / kg or 5 m