Question
Describe the preparation of $900 \mathrm{~mL}$ of $3.00 \mathrm{M} \mathrm{HNO}_{3}$ from the commercial reagent that is $70.5 \% \mathrm{HNO}_{3}$ (w/w) and has a specific gravity of $1.42$.
Step 1
Calculate the moles of HNO3 required for the desired solution: Moles = Molarity × Volume = 3.00 M × 0.900 L = 2.70 moles of HNO3 Show more…
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