[magicalatom]

Hi All,

Here is the problem:

What mass of H20 is need to prepare 148 g of 12% (m/m) KHCO3 solution?
A.  120 g
B. 130 g
C. 11.3 g
D. 17.8 g
E. 12.0 g

I calculated D.

Book is Answer B.

Book explanation is as follows:

% (m/m) means that both solute and solvent are measured by their mass.
If the solution contains 12% (m/m), it means that the rest of it is water, or:
100%-12% = 88%
Use this percentage to determine the mass:
mass of water = 88% * 148 grams
mass of water = 130.24 g aprrox. 130 grams


Now lol can someone explain this phenomenon here or was the author drunk/high?

I used % = is/of *100

[mikasaur]

The author was not drunk or high. And his or her explanation is a pretty good one.

Perhaps you did it backwards. By saying 12% they mean the solution is 12% solute, not solvent.

[magicalatom]

I think I sort of understand it now.

12% of 148 g = 17.76 grams

88% of 148 g = 130.24 grams

so i looked at it as (17.76/148) * 100 = 12%

The author was not drunk or high. And his or her explanation is a pretty good one.

Perhaps you did it backwards. By saying 12% they mean the solution is 12% solute, not solvent.

[mikasaur]

I think I sort of understand it now.

12% of 148 g = 17.76 grams

88% of 148 g = 130.24 grams

so i looked at it as (17.76/148) * 100 = 12%

Perhaps you just read the question wrong.

What mass of H20 is need to prepare 148 g of 12% (m/m) KHCO3 solution?

They're asking how much water is needed. Not how much potassium bicarbonate.