Question

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

O A. Put 120 grams of Mgo in the beaker and add exactly 1 L of water.
O B. Put 3 grams of Mgo in the beaker and add exactly 1 L of water.
O C. Put 3 grams of Mgo in the beaker and add enough water to reach the 1 L mark.
O D. Put 120 grams of Mgo in the beaker and add enough water to reach the 1 L mark.​

Answer 1

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

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

Answer 2

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

Step-by-step explanation:

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

`Molarity=(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}}=(xg)/(40g/mol)`

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

`3M=(x)/(40g/mol* 1L)`

`x=120g`

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