The answer is E. 5.0 mol NaCl in 3.0 L
The mass concentration (c) is the mass of the solute (m) divided by the volume of the mixture (V):
c = m/V
Now, let's go through all choices:
A. m = 175 g NaCl
V = 1.0 L
c = m/V = 175 g/1.0 L = 175 g/L
B. m = 58.5 g NaCl
V = 1.5 L
c = m/V = 58.5 g/1.5 L = 39 g/L
C. m = 29.3 g NaCl
V = 0.5 L
c = m/V = 29.3 g/0.5 L = 58.6 g/L
D. M = 2.5 mol NaCl
Since we have moles, and not grams, first we must calculate molar mass of NaCl which is the sum of atomic masses of the elements:
Mr(NaCl) = Ar(Na) + Ar(Cl) = 22.99 g + 35.45 g = 58.44 g
This means that in 1 mol there is 58.44g/L. So, in 2.5 mol there will be:
1 mol : 58.44 g/L = 2.5 mol : x
x = 2.5 * 58.44 g/L = 146.1 g/L
E. We will use the calculation from choice D.
Mr (NaCl) = 58.44 g.
In 1 mol there is 58.44g/L, so in 5 mol there will be:
1 mol : 58.44 g/L = 5 mol : x
x = 58.44 * 5 = 292.2 g/L
Therefore:
E > A > D > C > B
