This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.
In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:
[tex]M=\frac{n}{V}[/tex]
Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:
[tex]2.60g*\frac{1mol}{102.89g} =0.0253mol[/tex]
Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:
[tex]M=\frac{0.0253mol}{0.160L}=0.158M[/tex]
Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:
[tex]V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L[/tex]
That in milliliters turns out to be:
[tex]V=0.211L*\frac{1000mL}{1L}=211mL[/tex]
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