Calculating the Volume of a 40% Solution

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Discussion Overview

The discussion revolves around calculating the volume of a 40% NaNO3 solution that contains a specific amount of solute, with a focus on the implications of density and concentration definitions. Participants explore the relationships between mass, volume, and concentration in the context of a chemistry problem.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates that for a density of 1320 g/L, there are 320 grams of solute in one liter of solution, suggesting this corresponds to a 40% solution.
  • Another participant challenges this calculation, asserting that the correct volume for 12.5 grams of NaNO3 should be 24.1 mL and questions the logic behind the initial claim.
  • Participants discuss the ambiguity of whether the concentration is expressed as weight/weight (w/w) or weight/volume (w/v), noting that this distinction affects the calculations.
  • One participant identifies errors in the initial calculations, including the miscalculation of the mass of NaNO3 corresponding to 0.15 moles.
  • Several participants clarify that a 40% solution by weight means that 100 g of solution contains 40 g of solute, and they explore how to derive the mass of NaNO3 from the given density.
  • Another participant calculates that 528 grams of NaNO3 corresponds to 6.21 moles per liter, leading to a volume calculation of 24.1 mL for 0.15 moles of NaNO3.
  • Participants express uncertainty about how to determine the proportions of solute and solvent in the solution based on the density and concentration definitions.

Areas of Agreement / Disagreement

There is no consensus on the initial calculations, as participants present competing views on the correct volume of solution needed and the interpretation of concentration definitions. The discussion remains unresolved regarding the implications of density on the calculations.

Contextual Notes

Participants highlight limitations in the initial problem statement, including the lack of clarity on whether the concentration is w/w or w/v, and the assumptions made about the relationship between density and solution composition.

Who May Find This Useful

This discussion may be useful for students and individuals interested in chemistry, particularly those studying solution concentrations, density, and stoichiometry in the context of homework problems.

barryj
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Homework Statement
What volume of 40% NaNO3 solution contains 0.15 mole of NaNO3 ? Density = 1.32 g/mL = 1320 g/L
Relevant Equations
mwt of NaNO3 = 85 0.15 moles of NaNO3 = 12.5 grams
Soluability of NaNO3 is 900 g/L at 25 C
Since the density is 1320 g/L then for every L of solution there is 320 grams of solute.

320 is 35% of 900 hence the statement it is is a (about) a 40% solution

So the volume that would have 12.5 grams would be 12.5/320 L = .039 L = 39 mL
 
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The correct answer is 24.1 mL Where is my logic incorrect??
 
barryj said:
Homework Statement:: What volume of 40% NaNO3 solution contains 0.15 mole of NaNO3 ? Density = 1.32 g/mL = 1320 g/L
Relevant Equations:: mwt of NaNO3 = 85 0.15 moles of NaNO3 = 12.5 grams
Soluability of NaNO3 is 900 g/L at 25 C

Since the density is 1320 g/L then for every L of solution there is 320 grams of solute.

320 is 35% of 900 hence the statement it is is a (about) a 40% solution

So the volume that would have 12.5 grams would be 12.5/320 L = .039 L = 39 mL
The three problems I can spot are:

1. You haven’t stated if the concentration is 40% w/w or 40% w/v.

2. It is wrong to say “Since the density is 1320 g/L then for every L of solution there is 320 grams of solute.” (Though this may be approximately true).

If you dissolve 320g of a solute in 1 litre (1000g) of water, the total mass will be 1320g - but the volume will not still be 1L. So the density will not be 1320g/L.

3. 0.15 mol of NaNO₃ is not 12.5g
 
1. the stated problem didn't state w/w or w/v
2. I was afraid of this so this will make my solution incorrect
3. mwt of NaNO3 is 85. 0.15 * 85 = 12.75 I stated 12.5 is this the problem?

I did come up with this equation that gave the correct answer but I do not fully know why.

0.40 * V * 1.32 = 12.5
V = 24.1 ml the correct answer.

I do not totally undertand what a 40% solution is. 40% of what?
Maybe you could enlighten me.
 
40% w/w means that 100 g of solution contains 40 g solute.
40% w/v means that 100 ml of solution contains 40 g solute.
Let's assume that the solution is 40% w/w (that leads to the stated answer).
From the density, you know the mass of 1 L of solution. What mass of NaNO3 does this contain?
So what volume of solution would contain 12.75 g NaNO3?
 
A 40% solution of NaNO3 means that the solution is 40% NaNO3 by weight. If the density is 1320 g/L, this means that there are ##(0.4)(1320)## grams of NaNO3 per liter, or 528 grams NaNO3/l .

How many moles NaNO3/liter is that?

How many liters of this do you have to have for you to have 0.15 moles?
 
Chestermiller said:
A 40% solution of NaNO3 means that the solution is 40% NaNO3 by weight. If the density is 1320 g/L, this means that there are ##(0.4)(1320)## grams of NaNO3 per liter, or 528 grams NaNO3/l .

How many moles NaNO3/liter is that?

How many liters of this do you have to have for you to have 0.15 moles?
mwt of NaNO3 is 85. 528 grams would be 528/85 = 6.21 moles/liter
so, 6.21 m/1000mL = 0.15 m/x ml
solving gives x = 24.1 mL and this is correct.
mjc123 said:
40% w/w means that 100 g of solution contains 40 g solute.
40% w/v means that 100 ml of solution contains 40 g solute.
Let's assume that the solution is 40% w/w (that leads to the stated answer).
From the density, you know the mass of 1 L of solution. What mass of NaNO3 does this contain?
So what volume of solution would contain 12.75 g NaNO3?
From the density, you know the mass of 1 L of solution. What mass of NaNO3 does this contain? I guess I do not know how to find this. I would like to know how.
 
barryj said:
mwt of NaNO3 is 85. 528 grams would be 528/85 = 6.21 moles/liter
so, 6.21 m/1000mL = 0.15 m/x ml
solving gives x = 24.1 mL and this is correct.

From the density, you know the mass of 1 L of solution. What mass of NaNO3 does this contain? I guess I do not know how to find this. I would like to know how.
mjc123 said:
40% w/w means that 100 g of solution contains 40 g solute.
40% w/v means that 100 ml of solution contains 40 g solute.
Let's assume that the solution is 40% w/w (that leads to the stated answer).
From the density, you know the mass of 1 L of solution. What mass of NaNO3 does this contain?
So what volume of solution would contain 12.75 g NaNO3?
Assuming w/w 1 liter would be 1320 grams. How can you determine what part is water and what part is NaNO3.
 
barryj said:
Assuming w/w 1 liter would be 1320 grams. How can you determine what part is water and what part is NaNO3.
It's 40% by weight of NaNO3 and 60% water.
 
  • #10
barryj said:
Assuming w/w 1 liter would be 1320 grams. How can you determine what part is water and what part is NaNO3.
1 litre weighs 1320g because we are told this (the density). It is nothing to do with w/w or w/v is being used for concentration.

But using 40% w/w gives:
40% of 1320g = 528g
So 1 litre of solution contains:
528g of NaNO₃
1320g – 528g = 792g of water
 
  • #11
The ole brain has figured it out.
Thanks all.
 

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