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You have a 0.7 M solution. Your job is to produce 50 mL of a 0.4 M solution.


A. How much of the 0.7 M solution do you need to start with? (Show your work.)


B. How many moles of solute were in the 0.7 M solution? (Show your work.)


C. How much water do you need to add to the previous amount of the 0.7 M solution to dilute it to

0.4 M? (Show your work.)


D. A student decides to put double the amount of water calculated in Part C. Describe what effect this

will have on the overall concentration of the resulting solution. Justify your answer.

Sagot :

Answer: A. 28.6 ml of the 0.7 M solution we need to start with.

B. There are 0.02 moles of solute were in the 0.7 M solution.

C. Amount of water to be added is 21.4 ml

D. The resulting solution will have concentration of 0.28 M

Explanation:

According to the dilution law,

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of stock solution = 0.7 M

[tex]V_1[/tex] = volume of stock solution = ?

[tex]M_1[/tex] = molarity of diluted solution = 0.4 M

[tex]V_1[/tex] = volume of  diluted solution = 50 ml

Putting in the values we get:

[tex]0.7\times V_1=0.4\times 50[/tex]

[tex]V_1=28.6ml[/tex]

A. 28.6 ml of the 0.7 M solution we need to start with.

B. Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

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

[tex]0.7=\frac{n\times 1000}{28.6}[/tex]

[tex]n=0.02[/tex]

Thus there are 0.02 moles of solute were in the 0.7 M solution.

C. Amount of water to be added = (50-28.6 ) ml = 21.4 ml

D. If water added is [tex]2\times 21.4=42.8 ml[/tex]

[tex]0.7\times 28.6=M_2\times 71.4[/tex]

[tex]M_2=0.28M[/tex]

If volume of concentrated solution will be more , the resulting solution will have lesser concentration of 0.28 M

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