Question

How would you prepare a 1 L solution of 3 M Mgo?

Answer 1

Answer: We need to 120 g of MgO in the beaker and add enough water to reach the 1 L mark.​

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

moles of $MgO$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{40g/mol}$

Now put all the given values in the formula of molality, we get

$3M=\frac{x}{40g/mol\times 1L}$

$x=120g$

Therefore, we need to 120 g of MgO in the beaker and add  enough water to reach the 1 L mark.​