Answer: The volume for 0.850 mol of [tex]NaNO_{3}[/tex] from a [tex]NaNO_{3}[/tex] solution is 1700 mL.
The volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.
Explanation:
Molarity is the number of moles of solute present in a liter of solution.
- As given moles of [tex]NaNO_{3}[/tex] are 0.850 mol and molarity of [tex]NaNO_{3}[/tex] solution is 0.5 M. Hence, its volume is calculated as follows.
[tex]Molarity = \frac{no. of moles}{Volume (in L)}\\0.5 M = \frac{0.850 mol}{Volume}\\Volume = 1.7 L (1 L = 1000 mL)\\= 1700 mL[/tex]
Therefore, the volume for 0.850 mol of [tex]NaNO_{3}[/tex] from a [tex]NaNO_{3}[/tex] solution is 1700 mL.
- As given mass of LiOH are 30.0 g from a 2.70 M LiOH (molar mass = 23.95 g/mol) solution. Hence, its number of moles are calculated as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{30.0 g}{23.95 g/mol}\\= 1.25 mol[/tex]
So, volume for LiOH solution is calculated as follows.
[tex]Molarity = \frac{no. of moles}{Volume (in L)}\\2.70 M = \frac{1.25}{Volume}\\Volume = 0.464 L (1 L = 1000 mL)\\= 464 mL[/tex]
Therefore, volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.
