# What is the molality of 1M NH4OH in 70.092ml of water?

• HelloCthulhu
In summary: The Attempt at a SolutionSolve for molality:convert volume (ml) to mass (kg)70.092ml x 17.031g/ml = 864.291g / 1000 = 0.08641kg1 / 0.08641 = 14.3 molals/kgSolve for total mass:17.031g + 70.092g = 105.138gSolve for total volume:convert mass(g) to volume(ml)17.031g / 0.88
HelloCthulhu

## Homework Statement

What is the molality of 1M NH4OH dissolved in 70.092ml of water? What is the total mass of the solution? What is the total volume? What is the concentration percent? What is the molar concentration?

1M NH4OH = 35.046g
density of water = 1g/ml
density of NH4OH = 0.88 g/ml

## Homework Equations

molality = moles of solute/kg of solvent

mass = volume x density

molarity = moles of solute / liters of solution

concentration = moles / volume

## The Attempt at a Solution

Solve for molality:
convert volume (ml) to mass (kg)
70.092ml x 1g/ml =70.092g / 1000 = 0.07092kg
1 / 0.070092 = 14.3 molals/kg

Solve for total mass:
35.046g + 70.092g = 105.138g

Solve for total volume:
convert mass(g) to volume(ml)
35.046g / 0.88g/ml = 39.825ml
39.825ml + 70.092ml = 109.917ml

Solve for concentration percent:
35.046g / 109.917ml=0.3188 x 100 = 31.88406= 32%Solve for molar concentration:
1M / 0.109917L = 9.09777M/L = 9.1M/L

HelloCthulhu said:
What is the molality of 1M NH4OH dissolved in 70.092ml of water?

I am afraid this question doesn't make much sense. First, you can't dissolve "1 M" in anything, "1 M" is a concentration, result of already dissolving something. Chances are what you mean is "What is the molality of solution prepared by dissolving 1 mole of NH4OH in 70.092ml of water". But even then it is still wrong. Dissolving 1 mole of ammonia (NH3), would be OK. Diluting ammonia solution (which is sometimes thought of as it was solution of NH4OH), would be OK. But there is no way of dissolving NH4OH, there is no such compound that you can isolate.

orricl and HelloCthulhu
My mistake. I should have treated this as an ammonia solution. I'll rewrite it asking for the molality of 1 mole NH3 dissolved in 70.092ml of water.

Beware: the 0.88 g/ml density is also used incorrectly. It is density of the commercial solutions of ammonia (which are about 35%). To calculate concentrations precisely you need to find the density of your solution in tables.

You need to find the density of the solution you are working with. Not the density of the gas, not the density of the stock solution, these are not things you are working with.

Calculate the concentration (you have enough information to calculate at least molality and w/w %) and check the density of your solution in the tables.

Nothing wrong with not knowing and trying to learn so no need to apologize. IMHO wikihow gives a poor (in a pedagogical sense) example of using 3 mL of a powdered solid - that's not how you do it in reality (you measure volume of solutions, or you weigh solids), plus, it leaves people confused (just like you are).

Finding the density in tables is rather trivial. As I said - first, calculate the % w/w concentration of the SOLUTION (not a problem as you know mass of the solute and mass of the solvent). Then just look in the density tables, like this one:

http://www.chembuddy.com/?left=CASC&right=density_tables

or one that is more precise:

https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/density-of-ammonia/

(if there is no exact concentration that you have calculated just take the closest one, or estimate the needed value with a linear approximation). For diluted solutions you can assume their density to be just that of pure water, 1 g/mL.

Once you know the SOLUTION density (edit: and in this particular case the mass you have already calculated) you can calculate the SOLUTION volume, thus it will allow you to calculate volume based concentrations, like molarity or g/L (molality and w/w % don't require knowledge of the solution volume, as they are solely mass based).

Last edited:
HelloCthulhu
Borek said:
I am afraid this question doesn't make much sense. First, you can't dissolve "1 M" in anything, "1 M" is a concentration, result of already dissolving something. Chances are what you mean is "What is the molality of solution prepared by dissolving 1 mole of NH4OH in 70.092ml of water". But even then it is still wrong. Dissolving 1 mole of ammonia (NH3), would be OK. Diluting ammonia solution (which is sometimes thought of as it was solution of NH4OH), would be OK. But there is no way of dissolving NH4OH, there is no such compound that you can isolate.

Thank you so much for your help! I've reworked the problem now.

## Homework Statement

What is the molality of 1 mole of NH3 gas dissolved in 70.092ml of water? What is the total mass of the solution? What is the total volume? What is the concentration percent? What is the molar concentration?1mol NH3 = 17.031g

## Homework Equations

molality = moles of solute / kg of solventmass of solution = grams of solute + grams of solventmolarity = moles of solute / liters of solutionvolume = mass / densityconcentration = moles / volumemass percent of solution (w/w%) = mass of solute/mass of solution x 100

## The Attempt at a Solution

Solve for molality:

convert volume (ml) to mass (kg)

70.092ml x 1g/ml =70.092g / 1000 = 0.07092kg

1 / 0.070092 = 14.26 molals/kg

Solve for total mass:

17.031g + 70.092g = 87.123g

Solve for total volume:

Find w/w%

17.031g / 87.123g = 0.19548 x 100 = 19.548 = 20 w/w%Look up density table for 20 w/w% NH3 solution

density = 0.895187.123g / 0.8951g/ml = 97.333mlSolve for concentration %:

17.031g / 97.333ml = 0.1749 x 100 = 17.49 = 17.5 wt/vol %Solve for molar concentration:

1 mol / 0.097333L= 10.274M (mol/L)

Looks good

(Actually turned out there was a mistake in the density table at the chembuddy site, so you are a bit off in your numbers, but the reasoning is perfectly OK).

Last edited:
HelloCthulhu

## 1. What is molality and how is it different from molarity?

Molality is a measure of concentration that represents the number of moles of solute per kilogram of solvent. It is different from molarity, which represents the number of moles of solute per liter of solution. Molality takes into account the mass of the solvent, while molarity does not.

## 2. How do you calculate molality?

Molality (m) can be calculated by dividing the number of moles of solute (n) by the mass of the solvent (m) in kilograms. This can be expressed as m = n/m.

## 3. What is the significance of the units in the given molality question?

The units in the given question, 1M NH4OH in 70.092ml of water, are important because they determine the concentration of the solution. The M indicates molarity, while ml represents milliliters, which is a unit of volume. The use of ml instead of liters indicates that the solution is being measured in a small volume.

## 4. How do you convert from molarity to molality?

To convert from molarity (M) to molality (m), you need to know the density (ρ) of the solution. The equation for conversion is m = M/ρ, where ρ is in grams per milliliter (g/ml). This conversion is necessary because molarity is measured in moles per liter (mol/L) while molality is measured in moles per kilogram (mol/kg).

## 5. What is the molality of 1M NH4OH in 70.092ml of water?

The molality of 1M NH4OH in 70.092ml of water can be calculated by first converting the volume to kilograms. Since 1 milliliter of water has a mass of 1 gram, 70.092ml of water has a mass of 70.092 grams. We can then use the molality equation to calculate the molality: m = n/m = 1 mol / 0.070092 kg = 14.287 mol/kg. Therefore, the molality of 1M NH4OH in 70.092ml of water is 14.287 mol/kg.

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