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What mass (in g) of magnesium nitrate, Mg(NO3)2, are required to produce 250.0 mL of a 0.0750 M solution?

(A) 0.0188 g

(B) 0.0445 g

(C) 1.61 g

(D) 2.79 g


Solution:

Molarity = moles of solute/liters of solution

M = 0.0750M

V= 250.0ml( 1L/1000ml)= 0.250L


0.0750M = moles of solute/0.250L

moles of solute = 0.0188molMg(NO3)2

Now, lets convert moles to grams using the molar mass from the Periodic Table.

 0.0188molMg(NO3)2 ( 148.33g/1mol) = 2.79g Mg(NO3)2


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