Which of the following should be done in orde | vedclass

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(D) Initial moles of $NaCl$ in $100 \ mL$ of $0.30 \ M \ NaCl = \frac{100 	imes 0.3}{1000} = 0.03 \ mol$. Target moles of $NaCl$ in $100 \ mL$ of $0.

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$A$ and $C$ both

Vedclass · Answer

(D) Initial moles of $NaCl$ in $100 \ mL$ of $0.30 \ M \ NaCl = \frac{100 	imes 0.3}{1000} = 0.03 \ mol$. Target moles of $NaCl$ in $100 \ mL$ of $0.40 \ M \ NaCl = \frac{100 	imes 0.4}{1000} = 0.04 \ mol$. Moles of $NaCl$ to be added $= 0.04 \ mol - 0.03 \ mol = 0.01 \ mol$. Mass of $NaCl$ to be added $= 0.01 \ mol 	imes 58.5 \ g/mol = 0.585 \ g$. Since both adding $0.010 \ mol$ and adding $0.585 \ g$ of $NaCl$ result in the same outcome,both $A$ and $C$ are correct.

Which of the following should be done in order to prepare $0.40 \ M \ NaCl$ starting with $100 \ mL$ of $0.30 \ M \ NaCl$ (mol.wt. of $NaCl = 58.5$)?

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