Question

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

1. 116 g of 0.200 m NaNO3?

2. 116 g of 1.6 % NaNO3 solution by mass?

3. How much solvent would you take to prepare the solution in question 1?

4.How much solvent would you take to prepare the solution in question 2?

Answer 1

Q1:

Let mass of solute = x gm and mass of solvent would be (116-x)gm

Molality = mass of solute/molar mass × 1000/mass of solvent(in gm)

0.200 = (x/84.9947)×(1000/(116-x)) = 11.76544x/(116-x)

x = 1.9389 g

Mass of solute required = 1.94 g (answer)

Q2:

116 g of 1.6% NaNO3 by mass.

100g of solution has 1.6g of NaNO₃

116g of solution has (1.6/100)×116g = 1.856g

Mass of solute required = 1.856g = 1.86g. (Answer)

Q3:

Mass of solvent = mass of solution - mass of solute

= 116g - 1.9389g = 114.06g

= 114.1g. (Answer)

Q4:

Mass of solvent = 116g - 1.856g = 114.144g

= 114.14 g. (Answer)