Question

question 1 (2 points)
how many moles in 3.0 l of a 2.00 m solution of sodium chloride?
the conversion factor is always the concentration, in this case 1__
to calculate the moles we cancel out liters
2 mol
(3 l)(----------) = 4 mol
1 l
a. 0.00 b. 1.00 c. 2.00 d. 3.00 e. 2.00 mol/l
f. 4.00 g. 5.00 h. 6.00 i. 58.44 g/mol j. 58.44
k. 351 l. 10.00 m. 584.4

Explanation:

Step1: Identify the concentration

The concentration of the solution is given as \(2.00\space M\), which means \(2.00\space mol/L\). So, for blank 1, the conversion factor (concentration) is \(E. 2.00\space mol/L\).

Step2: Determine the value for moles in the conversion factor

In the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, and \(V\) is volume), the concentration \(C=\frac{n}{V}\). So, the number of moles in the conversion factor (numerator of the concentration fraction) for blank 2 is \(C. 2.00\).

Step3: Identify the volume value

The volume \(V\) of the solution is \(3.0\space L\). So, for blank 3, the volume value is \(D. 3.00\).

Step4: Calculate the number of moles

Using the formula \(n = C\times V\), substituting \(C = 2.00\space mol/L\) and \(V=3.0\space L\), we get \(n=(2.00\space mol/L)\times(3.0\space L)=6.0\space mol\). So, for blank 4, the number of moles is \(H. 6.00\).

Answer:

1. E. \(2.00\space mol/L\)
2. C. \(2.00\)
3. D. \(3.00\)
4. H. \(6.00\)