The concentration of commercially available concentrated nitric acid is 70.0 % b

Question

The concentration of commercially available concentrated nitric acid is 70.0 % by mass, or 15.9 M. Calculate the density and the molality of the solution. NH, is very soluble in water but NCI. is not. Explain clearly. Two beakers, one containing 50-ml of 1.0 M aqueous glucose solution and the other containing 50-ml of 2.0 M aqueous glucose solution, are place under a tightly sealed bell Jar at room temperature. What are the volumes in these two beakers at equilibrium? A solution is prepared by condensing 4.00 L of a gas, measured at 27degreeC and 748 mmHg pressure, into 58.0 g of benzene. Calculate the freezing point of this solution. (Sec Problem Set 18 for benzene data.)

Explanation / Answer

15.9 M solution means that 1 liter of solution contains 15.9 moles of HNO3.
Molar mass of HNO3 = 63 g /mol
Mass of HNO3 in 1 L solution: 15.9mol x 63 g/mol = 1007 g

Mass percentage of HNO3 in solution;

%HNO3 = (mass of HNO3 / total volume )*100

Assume that the total mass ix X ml

70 %HNO3 = (1007 / X) x 100

0.7 = 1007/X

X= 1438.6 ml

Density = mass/ volume

= 1007 g/ 1438.6 ml

= 1.43 g/ ml

Molality :number of moles / solvent mass in kg

Assume that the total mass of solution is 1000 g.

Now calculate the mass of HNO3 and water as follows:

(1000 g) x (710%) = 700 g HNO3

Now calculate the number of moles of HNO3 as follows:

(700 g HNO3) / (63.0 g HNO3/mol) = 11.11 mol HNO3

Solution mass – solute mass = solvent mass

1000 g solution - 700 g HNO3 = 300 g H2O or 0.300 kg

Molality = number of moles / mass of solute in kg

= 11.11 moles HNO3/ 0.300 kg

=37.0 m

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