What volume of a 0.33 M Mg(NO3)2 solution contains 96 g of Mg(NO3)2?
To find the volume of the solution that contains 96 g of Mg(NO3)2, we can use the formula:
Volume (in liters) = mass (in grams) / molar mass (in g/mol) / concentration (in mol/L)
First, let's calculate the molar mass of Mg(NO3)2:
Molar mass of Mg(NO3)2 = (1 * molar mass of Mg) + (2 * molar mass of NO3)
= (1 * 24.31 g/mol) + (2 * ((1 * 14.01 g/mol) + (3 * 16.00 g/mol)))
= 24.31 g/mol + (2 * (14.01 g/mol + 48.00 g/mol))
= 24.31 g/mol + (2 * 62.01 g/mol)
= 24.31 g/mol + 124.02 g/mol
= 148.33 g/mol
Now, let's substitute the values into the formula:
Volume = 96 g / 148.33 g/mol / 0.33 mol/L
Divide 96 g by 148.33 g/mol to get the number of moles:
Volume = 0.6465 mol / 0.33 mol/L
Finally, divide the number of moles by the concentration to get the volume:
Volume = 1.96 L
Therefore, the volume of a 0.33 M Mg(NO3)2 solution that contains 96 g of Mg(NO3)2 is 1.96 liters.