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What mass (in grams) of NH4Cl is needed to prepare 664 mL of 4.423 M NH CI solution?

Sagot :

Answer:

157.096grams

Explanations:

The formula for calculating the molarity of a solution is expressed as:

[tex]molarity=\frac{mole}{volume}[/tex]

Given the following parameters

[tex]\begin{gathered} volume=664mL=0.664L \\ molarity=4.423M \end{gathered}[/tex]

Determine the mole of the solution

[tex]\begin{gathered} mole=molarity\times volume \\ mole=4.423\times0.664 \\ mole=2.937moles \end{gathered}[/tex]

Determine the mass of NH4Cl

[tex]\begin{gathered} mass\text{ }of\text{ NH}_4Cl=mole\times molar\text{ mass} \\ mass\text{ }of\text{ NH}_4Cl=2.937\times53.491 \\ mass\text{ }of\text{ NH}_4Cl=157.096grams \end{gathered}[/tex]

Hence the mass in grams of ammonium chloride is 157.096grams

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