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What is the molality of ammonia in a solution containing 0.85 gram of NH3 in hundred centimetre cube of a liquid of density 0.85 gram per centimetre cube?
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What is the molality of ammonia in a solution containing 0.85 gram of ...
Explanation:

Molality is defined as the number of moles of solute present in 1000 grams (1 kg) of solvent. It is denoted by m.

Given:

Mass of NH3 = 0.85 g

Volume of liquid = 100 cm3

Density of liquid = 0.85 g/cm3

Solution:

First, we need to calculate the number of moles of NH3 present in the given mass.

Molar mass of NH3 = 14 + 3 = 17 g/mol

Number of moles of NH3 = Mass of NH3 / Molar mass of NH3

= 0.85 g / 17 g/mol

= 0.05 mol

Now, we need to calculate the mass of the solvent in the given volume.

Mass of liquid = Volume of liquid x Density of liquid

= 100 cm3 x 0.85 g/cm3

= 85 g

Finally, we can calculate the molality of the solution using the formula:

Molality (m) = Number of moles of solute / Mass of solvent (in kg)

= 0.05 mol / 0.085 kg

= 0.59 mol/kg

Therefore, the molality of ammonia in the given solution is 0.59 mol/kg.

Answer:

The molality of ammonia in the given solution is 0.59 mol/kg.
Community Answer
What is the molality of ammonia in a solution containing 0.85 gram of ...
M=moles/kg=0.85÷17/0.85*100*10^-3=10/17=0.58
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What is the molality of ammonia in a solution containing 0.85 gram of NH3 in hundred centimetre cube of a liquid of density 0.85 gram per centimetre cube?
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