Can someone help me in this? Concentrated HNO_3 has specific gravity of 1.42. It contains 69% w/w (weight by weight) of HNO_3. Calculate the molarity and molality of the solution?

Jul 8, 2015

Here's how you can solve for the molarity and molality of the solution.

Explanation:

A liquid's specific gravity is simply the ratio between its density and the density of a reference liquid, more often than not water, at the same temperature.

Usually, the specific gravity for liquids is given at ${4}^{\circ} \text{C}$, the temperature at which water's density is at its maximum value of approximately 1 g/mL.

So, in other words, using the specific gravity of a concentrated nitric acid solution is simply another way of saying that its density is 1.42 g/mL.

$\text{S.G" = rho_(HNO_3)/rho_"water" => rho_(HNO_3) = "S.G." * rho_"water}$

${\rho}_{H N {O}_{3}} = 1.42 \cdot \text{1 g/mL" = "1.42 g/mL}$

Now for the actual calculations. Select a 1.0-L sample of your stock solution, then use its density to determine what its mass is.

1.0cancel("L") * (1000cancel("mL"))/(1cancel("L")) * "1.42 g"/(1cancel("mL")) = "1420 g"

You also know the solution's percent concentration by mass to be 69%. This of cours implies that every 100 g of solution will contain 69 g of nitric acid.

The mass of nitric acid your sample contains will be

1420cancel("g solution") * ("69 g "HNO_3)/(100cancel("g solution")) = "979.8 g" $H N {O}_{3}$

Use nitric acid's molar mass to determine the number of moles you have

979.8cancel("g") * "1 mole"/(63.01cancel("g")) = "15.55 moles" $H N {O}_{3}$

The solution's molarity, which is defined as the number of moles of solute, in your case nitric acid, divided by the volume of the solution - in liters - will thus be

C = n/V = "15.55 moles"/"1.0 L" = color(green)("15.6 M")

To get the solution's molality, which is defined s moles of solute per kilograms of solvent, you need to first figure out how much water you have.

${m}_{\text{sol" = m_"nitric acid" + m_"water}}$

${m}_{\text{water" = 1420 - 979.8 = "440.2 g water}}$

Therefore, the molality will be

b = n_"nitric acid"/m_"water" = "15.55 moles"/(440.2 * 10^(-3)"kg") = color(green)("35,3 molal")

SIDE NOTE I left the answers rounded to three sig figs.