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

Mass = 889.8 g

Explanation:

Given dа ta:

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