Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?

A. 132 mL of 0.100 M NaNO3

B. 126 g of 0.220 m NaNO3

C. 126 g of 2.0 % NaNO3 solution by mass

D. How much solvent would you take to prepare the solution in part B? Part C?

Answer 1

First of all you need to know what is M and m and mass percent here.

M = Molarity = number of moles of solute per litre of solution (and not the solvent)

m = Molality = number of moles of solute per kilogram of solvent.

Mass Percent = grams of solute divided by gram of solution when multiplied by 100 gives the concentration of the solution in mass percent.

Secondly you need the molecular weight of the solute, which is 84.9947 gm/mol here for NaNO3.

A.  

1000 ml 1 molar(M) solution needs 84.9947 gm of NaNO3

132 ml 0.100 M solution will need (84.9947 x 132 x 0.1) / 1000 = 1.12193 gm of NaNO3.

B.

(1000+84.9947) = 1084.9947 gm of 1 m solution needs 84.8847 gm of NaNO3.

126 gm of 0.220 m solution will need (84.8847 x 126 x 0.220) / 1084.9947 = 2.17066779 gm of NaNO3.

C.

By formula,   

Therefore, grams of solute needed = (2 x 126) / 100 = 2.52 grams.

D.

For part B, we need (126 - 2.17066779) gm = 123.829332 gm of solvent

For part C, we need (126 - 2.52) = 123.48 gm of solvent.

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