Home › AP Chemistry › Chemistry › Composition of Substances and Solutions › A 0.150 L sample of 0.556 M NaCl is left to evap…
A 0.150 L sample of 0.556 M NaCl is left to evaporate until only 0.105 L remains. What is the new concentration?
A0.556 M
B0.389 M
C0.794 M
D0.0834 M
Answer & Solution
Correct answer: C. 0.794 M
1. Evaporation removes solvent only, so the amount of NaCl in the flask never changes.
2. That amount is 0.556 mol/L x 0.150 L = 0.0834 mol of NaCl.
3. The same moles now sit in a smaller volume of 0.105 L.
4. The new concentration is 0.0834 mol / 0.105 L = 0.794 mol per litre.
5. Losing volume with fixed solute must raise the concentration, so any answer below 0.556 M fails the sanity check.
6. The value 0.389 M applies the volume ratio the wrong way round, as though the solution had been diluted.
7. The value 0.556 M assumes concentration cannot change, and 0.0834 M reports the mole count as a molarity.
_Source: OpenStax Chemistry (CC BY 4.0), Ch 3 "Composition of Substances and Solutions", section 3.3 Molarity_
Related questions
In ammonia, NH3, the three hydrogen atoms together weigh 3.024 amu and the formula mass isWhat is the mass of 0.600 mol of ozone molecules, O3? Use O = 16.00 g/mol.Aspirin is C9H8O4 with a molar mass of 180.159 g/mol. Using H = 1.008 g/mol, what is its mThe EPA maximum safe level of fluoride ion in tap water is 4 ppm. At that level, what massLiquid bleach is 7.4% sodium hypochlorite by mass. What mass of NaOCl is in a 250.0 g bottD5W intravenous fluid is 5.0% dextrose, C6H12O6, by mass and has a density of 1.029 g/mL. If 4.12 L of 0.850 M H3PO4 is diluted to a final volume of 10.00 L, what is the resulting What volume of a 1.00 M Fe(NO3)3 stock can be diluted to prepare 1.00 L of a 0.250 M solut