Question

134 grams of nitric acid is added to 512 grams of water. Calculate the molality of nitric acid.

Answer 1

Answer: 4.15234 m

512 g H2O *
`(1 kg)/(1000 g)` = 0.512 kg H2O

Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol

H = 1.008 g/mol

N = 14.007 g/mol

O3 = 3*15.999

134 g HNO₃ *
`(mol)/(63.012 g)` = 2.126 mol

m =
`(2.126 mol)/(0.512 kg)` = 4.15234 m

Answer 2

0.004 m

Step-by-step explanation:

Given data:

Mass of nitric acid = 132 g

Mass of water = 512 g

Molality of nitric acid = ?

Solution:

Formula:

Molality = number of moles of solute / mass of solvent in Kg

Number of moles of nitric acid:

Number of moles = mass/molar mass

Number of moles = 132 g/ 63.01 g/mol

Number of moles = 2.09 mol

Molality:

m = 2.09 mol / 512 Kg

m = 0.004 m