Question

Note: (show all work, and use atomic weights rounded to 2 decimal places). What volume of a 1.63 mass % solution of Mg(NO₃)_{2 ​}is needed to contain 0.316 grams of Mg(NO₃)₂​? The solution has a density of 1.016 g/ml. _________

Answer 1

Let the mass of solution [Water + MgNO₃)₂]= 'A ' g.

Mass of MgNO₃)₂ = 0.316 g

Also give that its a 1.63 (%w/w) solution.

i.e. (%w/w) solution = [(Mass of MgNO₃)₂ / Mass of solution] x 100

Using  given values,

1.63 = [0.316 / A] x 100

1.63 = 31.6 / A

A = 31.6 / 1.63

A = 19.39 g.

Now,

Mass Mg(NO₃)₂ of solution = 19.39 g

Density of solution = 1.016 g/mL

Volume of Mg(NO₃)₂ solution = ?

Formula,

Density = Mass / Volume

Volume = Mass / Density

= 19.39 / 1.016

= 19.08 mL

Volume of aq.Mg(NO₃)₂ solution is 19.08 mL.

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