Question

How many moles of Ba(NO3)2 are there in 0.25 L of a 2.00 M Ba(NO3)2 solution? 0.13 mol 0.50 mol 2.25 mol 8.0 mol

Answer 1

B. 0.50 mol

Step-by-step explanation:

Answer 2

Answer : The number of moles present in barium nitrate solution is, 0.50 moles

Solution : Given,

Molarity of barium nitrate solution = 2 M = 2 mole/L

Volume of solution = 0.25 L

Molarity : Molarity is defined as the number of moles of solute present in one liter of solution.

Formula used :

`Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution in liter}}`

or,

`Molarity=\frac{\text{Moles of }Ba(NO_3)_2}{\text{Volume of solution in liter}}`

Now put all the given values in this formula, we get the moles of barium nitrate.

`2mole/L=\frac{\text{Moles of }Ba(NO_3)_2}{0.25L}`

`\text{Moles of }Ba(NO_3)_2=0.50moles`

Therefore, the number of moles present in barium nitrate solution is, 0.50 moles