Answer:
The molarity of the [tex]\rm NaCl[/tex] solution in the first beaker would continue to be [tex]0.700\; \rm M[/tex].
Explanation:
Assume that the volume of the original solution is [tex]v_{0}\; \rm L[/tex], and that [tex]v_{1}\; \rm L[/tex] of the solution was transferred to the other beaker. (Because of the [tex]\rm NaCl[/tex] in the solution, [tex]v[/tex] might not necessarily be [tex]50.0\; \rm mL[/tex].) In this question, [tex]v_{0} \ne v_{1}[/tex].
The molarity of [tex]\rm NaCl[/tex] in the solution of the first beaker was [tex]0.700\; \rm M[/tex] before the transfer. Calculate the number of moles of [tex]\rm NaCl\![/tex] in that solution:
[tex]\begin{aligned} & n(\text{original}) \\ =\; & c \cdot V \\ =\; & 0.700\; {\rm mol} \cdot L^{-1} \cdot (v_{0}\; {\rm L}) \\ =\; & 0.700\, v_{0}\; \rm mol \end{aligned}[/tex].
The solution in the first beaker is uniform, such that the molarity of that [tex]v\; \rm L[/tex] of transferred solution would be [tex]0.700\; \rm M[/tex]. Calculate the number of moles of [tex]\rm NaCl[/tex] transferred to the other beaker:
[tex]\begin{aligned} & n(\text{transferred}) \\ =\; & c \cdot V \\ =\; & 0.700\; {\rm mol} \cdot L^{-1} \cdot (v_{1}\; {\rm L}) \\ =\; & 0.700\, v_{1}\; \rm mol \end{aligned}[/tex].
After the transfer, the number of moles of [tex]\rm NaCl[/tex] in the first beaker would be:
[tex]\begin{aligned}n &= n(\text{original}) - n(\text{transferred}) \\ &= 0.700\, v_{0}\; {\rm mol} - 0.700\, v_{1}\; {\rm mol} \\ &= 0.700\, (v_{0} - v_{1})\; {\rm mol}\end{aligned}[/tex].
After the transfer, the volume of the solution in the first beaker would be [tex](v_{0} - v_{1})\; \rm L[/tex].
Calculate the molarity of the solution in the first beaker after the transfer:
[tex]\begin{aligned} c &= \frac{n}{V} \\ &= \frac{0.700\, (v_{0} - v_{1})\; {\rm mol}}{(v_{0} - v_{1})\; {\rm L}} \\ &= 0.700\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
In other words, moving part of a solution from one container to another would not affect the molarity of the original solution.
