Answer:
0.56M NaCl is the final solution concentration.
Step-by-step explanation:
1.5 M NaCl
Change the M to its actual unit: 1.5 moles/liter [M means moles/liter]
Next, calculate the moles of NaCl contained in 185 ml of 1.5 moles/liter NaCl solution. Use the units to guide the calculation. A conversion factor is needed since one measure is in ml and the other in L. We know that (1 liter/1000 ml) is a conversion factor suitable for this.
(1.5 moles/liter NaCl)*(185 ml NaOH)*(1 liter/1000 ml) = 0.2775 moles NaCl
[The liter and ml units both cancel, leaving just moles]
This number of moles of NaCl (in the 185 ml) is then diluted to 0.5 L, That gives us a new concentration of (0.2775 moles NaCl)/(0.5 L).
The result (0.2775/0.5)(moles/L) is 0.555 moles/L or 0.56 M, to 2 sig figs.
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Another approach is using the relationship M1V1 = M2V2, where M and V are the concentration and volume, respectively. The states 1 and 2 represent the original (1) and final (2) results.
We should get the same answer as before, so it makes a good check.
M1V1 = M2V2
M1 = 1.5M
V1 = 185 ml
M2 = Unknown
V2 = 500 ml (0.5L)
M1V1 = M2V2
M2 = (M1*V1)/V2
M2 = (1.5M)*(185ml)/(500ml)
M2 = (1.5M)*(0.370)
M2 = 0.555M, or 0.56M
This is the same value as we derived at the start. Two ways of achieving the same result. Use the units to help guide the arithmetic operations, and to decide whether conversion factors are necessary.