Answer:
222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.
Explanation:
Being:
the molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:
Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole
So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?
[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]
amount of moles=0.0289 moles
Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:
[tex]molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in [tex]\frac{moles}{liter}[/tex]
So in this case:
Replacing:
[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]
Solving:
[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]
volume=0.2223 liters
Being 1 L= 1,000 mL:
volume=0.222 liters= 222.3 mL
222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.
