Question

How many grams of magnesium nitrate (Mg(NO3)2) are required to make a 4M solution in 1.5 L of solution

Answer 1

Answer:

Mass = 889.8 g

Explanation:

Given data:

Mass of Mg(NO₃)₂ required = ?

Volume of solution = 1.5 L

Molarity of solution = 4 M

Solution:

we will calculate the number of moles from molarity formula then we will calculate the mass from number of moles.

Formula:

Molarity = number of moles / Volume in L

4 M =  number of moles / 1.5 L

Number of moles  = 4 M× 1.5 L

Number of moles  = 6 mol/L × L

Number of moles  = 6 mol

Mass of Mg(NO₃)₂:

Mass= number of moles ×molar mass

Mass =  6 mol × 148.3 g/mol

Mass = 889.8 g