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

Question

asked Aug 24, 2019 235k views

2 Answers

JPTremblay · Answer 1 · 2019-08-28T15:01:02+0000

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