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.​

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.

where,

n = moles of solute

= volume of solution in L

moles of =

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

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

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

pie chart because it is in percents ( % )