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- When calculating the heat of solution, you should include the solute mass with your solution mass. I would appreciate some clarification regarding some discrepancies in my understanding of this.

When solving "coffee cup calorimeter" problems, you're supposed to include the solute mass with the mass of your solution.

However, you're also supposed to assume that dilute solutions have the same density and heat capacity as water.

So if I add 5g of NaOH to 500g of water, the solution volume likely remains 500 mL (due to negligible volume change).

When solving for the heat of solution in this case (plugging into Q=mct), should I use 505g as the mass of the solution (solute+solvent) or 500g (density of dilute solution is assumed to be 1g/mL, same as water at room temp)?

Furthermore, when adding solute mass to solvent mass, wouldn't you end up over-estimating the heat capacity of the solution?

Water has a fairly high heat capacity of 4.18 J/g C. Meanwhile, I assume the heat capacity of NaOH is significantly lower.

Wouldn't including the extra 5g of solute, and assuming it's heat capacity is the same as water give you a higher value than it should be?

Thanks in advance for clarifying!

However, you're also supposed to assume that dilute solutions have the same density and heat capacity as water.

So if I add 5g of NaOH to 500g of water, the solution volume likely remains 500 mL (due to negligible volume change).

When solving for the heat of solution in this case (plugging into Q=mct), should I use 505g as the mass of the solution (solute+solvent) or 500g (density of dilute solution is assumed to be 1g/mL, same as water at room temp)?

Furthermore, when adding solute mass to solvent mass, wouldn't you end up over-estimating the heat capacity of the solution?

Water has a fairly high heat capacity of 4.18 J/g C. Meanwhile, I assume the heat capacity of NaOH is significantly lower.

Wouldn't including the extra 5g of solute, and assuming it's heat capacity is the same as water give you a higher value than it should be?

Thanks in advance for clarifying!