# Question 7444f

May 11, 2015

SIDE NOTE This problem can't be solved without the density of the potassium nitrate solution.

Now, an educated guess would have me say that the density of this solution was probably given to as as being 1.02 - 1.08 g/mL, so I'll pick a density of 1.05 g/mL.

If this was not the value given to you, redo all the calculations with whatever value you have.

Molality is defined as moles of solute, in your case potassium nitrate, per kilogram of solvent.

This means that you have to use the density of the solution to determine how many grams of water you have.

$\rho = \frac{m}{V} \implies m = \rho \cdot V$

${m}_{\text{sol" = 1.05"g"/cancel("mL") * (1000cancel("mL"))/(1cancel("L")) * 1.89cancel("L") = "1984.5 g}}$

Out of this, 79.5 g will be potassium nitrate, which means that the mass of water will be

${m}_{\text{water" = m_"sol}} - {m}_{K N {O}_{3}}$

${m}_{\text{water" = 1984.5 - 79.5 = "1905 g}}$

Use potassium nitrate's molar mass to determine how many moles you have present in solution

79.5cancel("g") * ("1 mole "KNO_3)/(101.1cancel("g")) = "0.7864 moles" $K N {O}_{3}$

The solutiuon's molality will thus be

b = "0.7864 moles"/(1905 * 10^(-3)"kg") = color(green)("0.413 molal")

The percent concentration by mass is calculated as mass of solute divided by mass of solution and multiplied by 100.

"%m/m" = m_(KNO_3)/(m_"solution") * 100

"%m/m" = (79.5cancel("g"))/(1984.5cancel("g")) * 100 = color(green)("4.01%")#