Question

question 2 (2 points)
how many grans in 3.0 l of a 2.00 m solution of sodium chloride if the gfw is 58.44 g?
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
to get the grams of sodium chloride we use the molar mass, 5
6 g
(7 mol)(---------) = 8 g
1 mol
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: Determine the conversion factor (concentration)

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

Step2: Calculate the moles of \(NaCl\)

We use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, and \(V\) is volume). The volume \(V = 3.0\space L\) and \(C=2.00\space mol/L\).
\((3.0\space L)\times(\frac{2.00\space mol}{1\space L})\)
For blank 2, it's \(2.00\) (the numerator of the concentration fraction), for blank 3, it's \(3.0\) (the volume), and for blank 4, \(n=(3.0)\times(2.00)=6.00\space mol\) (so blank 4 is \(H. 6.00\)).

Step3: Calculate the mass of \(NaCl\)

We use the formula \(m=n\times M\) (where \(m\) is mass, \(n\) is moles, and \(M\) is molar mass). The molar mass \(M = 58.44\space g/mol\) (blank 5 is \(I. 58.44\space g/mol\), blank 6 is \(58.44\)). We found \(n = 6.00\space mol\) (blank 7 is \(H. 6.00\)).
\(m=(6.00\space mol)\times(\frac{58.44\space g}{1\space mol})\)
\(m = 6.00\times58.44=350.64\approx351\space g\) (blank 8 is \(K. 351\)).

Answer:

1. E. \(2.00\space mol/L\)
2. C. \(2.00\)
3. D. \(3.00\)
4. H. \(6.00\)
5. I. \(58.44\space g/mol\)
6. J. \(58.44\)
7. H. \(6.00\)
8. K. \(351\)