Question

If 0.38 M of an aqueous solution of magnesium nitrate hexahydrate (Mg(NO3)2 ᐧ 6H2O) was prepared in the laboratory, how many mL of the solution would you need to obtain 4.871 grams of this solute?

Answer 1

Answer:

50 mL

Explanation:

Given data

• Molar concentration (M): 0.38 M

• Mass of solute: 4.871 g

• Molar mass of the solute: 256.41 g/mol

We can find the required volume using the following expression.

M = mass of solute / molar mass of solute × liters of solution

liters of solution = mass of solute / molar mass of solute × M

liters of solution = 4.871 g / 256.41 g/mol × 0.38 mol/L

liters of solution = 0.050 L = 50 mL

Answer 2

Answer:

The correct answer is 49.994 ml (≅ 50 ml)

Explanation:

From the definition of molarity, we know that in a 0.38 M solution there are 0.38 mol of magnesium nitrate hexahydrate per liter of solution:

0.38 M= 0.38 mol / L

The first step to solve the problem is to convert the moles of solute to mass (in grams) by using the molecular weight of Mg(NO₃)₂.6H₂O.

Molecular weight of Mg(NO₃)₂.6H₂O= 256.41 g/mol (from the molar mass of the elements).

mass of Mg(NO₃)₂.6H₂O = 0.38 mol x 256.41 g/mol= 97.43 g

So, there are 97.43 g of Mg(NO₃)₂.6H₂O in 1 L of solution (1 L= 1000 ml). We need to know in how many ml of this solution a mass of 4.871 g is contained, so we write the relations and calculate x (in ml):

97.43 g Mg(NO₃)₂.6H₂O --------------------  1000 ml (= 1 L)

4.871 g Mg(NO₃)₂.6H₂O --------------------- x

x= 4.871 g x 1000 ml / 97.43 g= 49.994 ml