Question

What mass of strontium nitrate (Sr(NO3)2) would be required to prepare 2.000 L of a 0.0150 M aqueous solution of this salt?

Answer 1

To prepare a 0.0150 M solution of strontium nitrate in 2.000 L, you would need to use 4.49 grams of strontium nitrate.

Step-by-step explanation:

To calculate the mass of strontium nitrate required to prepare the solution, we can use the formula:

moles = concentration x volume

First, convert the concentration to moles per liter:

moles/liter = concentration = 0.015 M

Next, multiply the moles per liter by the volume in liters to find the moles of strontium nitrate:

moles = (0.015 M) x (2.000 L) = 0.030 mol

Finally, calculate the mass of strontium nitrate using the molar mass:

mass = moles x molar mass = 0.030 mol x 149.6 g/mol = 4.49 g

Answer 2

Answer : The mass of strontium nitrate required would be, 6.35 grams.

Explanation : Given,

Volume of solution = 2.000 L

Molar mass of
`Sr(NO_3)_2` = 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 :

`\text{Molarity}=\frac{\text{Mass of }Sr(NO_3)_2}{\text{Molar mass of }Sr(NO_3)_2* \text{Volume of solution (in L)}}`

Now put all the given values in this formula, we get:

`0.0150M=\frac{\text{Mass of }Sr(NO_3)_2}{211.63g/mole* 2.000L}`

`\text{Mass of }Sr(NO_3)_2=6.35g`

Therefore, the mass of strontium nitrate required would be, 6.35 grams.