Meghann321idn
Answered

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a.)What volume of 8.40 N Al(OH)3 will be used to prepare 350.0 mL of 2.10 M solution?

b.)How many mL of water will be used to prepare this dilute solution? ​

Sagot :

Tokamakidn

[tex]\tt{\huge{\red{a) \: Solution:}}}[/tex]

Step 1: List the given values.

Note that Al(OH)₃ has three replaceable OH⁻ ions. This is the reason why the given normality must be divided by 3 to convert from normality to molarity.

[tex]\begin{aligned} & M_{\text{i}} = \frac{8.40 \: N}{3} = 2.80 \: M \\ & M_{\text{f}} = 2.10 \: M \\ & V_{\text{f}} = \text{350.0 mL} \end{aligned}[/tex]

Final Step: Calculate the volume of the stock solution.

[tex]\begin{aligned} V_{\text{i}} & = \frac{M_{\text{f}}V_{\text{f}}}{M_{\text{i}}} \\ & = \frac{(2.10 \: M)(\text{350.0 mL})}{2.80 \: M} \\ & = \boxed{\text{262.5 mL}} \end{aligned}[/tex]

Hence, the volume of 8.40 N Al(OH)₃ solution is 262.5 mL.

[tex]\tt{\huge{\red{b) \: Solution:}}}[/tex]

[tex]\begin{aligned} V_{\text{water}} & = V_{\text{f}} - V_{\text{i}} \\ & = \text{350.0 mL} - \text{262.5 mL} \\ & = \boxed{\text{87.5 mL}} \end{aligned}[/tex]

Hence, the volume of water will be used is 87.5 mL.

[tex]\\[/tex]

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