Question

How many mL of a 0.130 M aqueous solution of chromium(II) nitrate, Cr(NO3)2, must be taken to obtain 5.08 grams of the salt

Answer 1

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.

Step-by-step explanation:

Being:

• Cr: 52 g/mole
• N: 14 g/mole
• O: 16 g/mole

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?

`amount of moles=(5.08 grams* 1 mole)/(176 grams)`

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:

`molarity (M)=(number of moles of solute)/(volume)`

Molarity is expressed in
`(moles)/(liter)`

So in this case:

• molarity= 0.130 M
• number of moles of solute= 0.0289 moles
• volume= ?

Replacing:

`0.130 M=0.130 (moles)/(liter) =(0.0289 moles)/(volume)`

Solving:

`volume=(0.0289 moles)/(0.130 (moles)/(liter) )`

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.