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A 72.84 g sample of Ba(OH)_2Ba(OH) 2 ​ is dissolved in enough water to make 1.700 liters of solution. How many mL of this solution must be diluted with water in order to make 1.000 L of 0.100 M Ba(OH)_2Ba(OH) 2 ​

Respuesta :

Answer : 399.8 ml of [tex]Ba(OH)_2[/tex] solution must be diluted with water.

Solution : Given,

Mass of [tex]Ba(OH)_2[/tex] = 72.84 g

Volume of solution = 1.7 L

Molarity of water = 0.1 mole/L

Volume of water = 1 L

First we have to calculate the molarity of [tex]Ba(OH)_2[/tex] solution.

Formula used : [tex]M=\frac{w_b}{M_b\times V_s}[/tex]

where,

M = molarity of [tex]Ba(OH)_2[/tex] solution

[tex]w_b[/tex] = mass of [tex]Ba(OH)_2[/tex]

[tex]M_b[/tex] = molar mass of [tex]Ba(OH)_2[/tex]

[tex]V_s[/tex] = volume of solution

Now put all the given values in this formula, we get

[tex]M=\frac{72.84g}{(171.34g/mole)\times (1.7L)}[/tex]

[tex]M=0.25007mole/L[/tex]

Now we have to calculate the volume of [tex]Ba(OH)_2[/tex] solution.

Formula used : [tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]Ba(OH)_2[/tex] solution

[tex]M_2[/tex] = molarity of water

[tex]V_1[/tex] = volume of [tex]Ba(OH)_2[/tex] solution

[tex]V_2[/tex] = volume of water

Now put all the given values in this formula, we get

[tex](0.25007mole/L)V_1=(0.1mole/L)\times (1L)[/tex]

By rearranging the term, we get

[tex]V_1=0.3998L=399.8ml[/tex]       (1 L = 1000 ml)

Therefore, 399.8 ml of [tex]Ba(OH)_2[/tex] solution must be diluted with water.

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