Answer : The mass of strontium nitrate required would be, 6.35 grams.
Explanation : Given,
Volume of solution = 2.000 L
Molar mass of [tex]Sr(NO_3)_2[/tex] = 211.63 g/mole
Molarity = 0.0150 M
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Mass of }Sr(NO_3)_2}{\text{Molar mass of }Sr(NO_3)_2\times \text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]0.0150M=\frac{\text{Mass of }Sr(NO_3)_2}{211.63g/mole\times 2.000L}[/tex]
[tex]\text{Mass of }Sr(NO_3)_2=6.35g[/tex]
Therefore, the mass of strontium nitrate required would be, 6.35 grams.
