Question

question 4 (3 points)
how would you make 5.00 l of a 2.00 m solution of sodium chloride if the gfw is 58.44 g?
to calculate the moles we cancel out liters
1 mol
(2 l)(----------) = 3 mol
1 l
to get the grams of sodium chloride we use the molar mass, 4
5 g
(6 mol)(---------) = 7 g
1 mol
we dissolve the nacl into water to avoid any changes in temperature and volume, then add water so that the bottom of the meniscus is on top of the 5.00 l mark, then stir to make homogeneous.
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: Calculate moles of NaCl

The formula for molarity \(M=\frac{n}{V}\) (where \(n\) is moles and \(V\) is volume in liters). Rearranged for \(n\), we have \(n = M\times V\). Here \(M = 2.00\space mol/L\) (the value for blank 1 and the denominator in the mole - calculation formula), \(V=5.00\space L\) (blank 2).

$$n=(5.00\space L)\times\frac{2.00\space mol}{1\space L}=10.00\space mol$$

(blank 3)

Step2: Calculate mass of NaCl

The formula for mass \(m=n\times M_m\) (where \(n\) is moles and \(M_m\) is molar mass). The molar mass of \(NaCl\) is \(58.44\space g/mol\) (blank 4 and the denominator in the mass - calculation formula). We found \(n = 10.00\space mol\) (blank 6).

$$m=(10.00\space mol)\times\frac{58.44\space g}{1\space mol}=584.4\space g$$

(blank 7)

Answer:

1. C. \(2.00\)
2. G. \(5.00\)
3. L. \(10.00\)
4. I. \(58.44\space g/mol\)
5. J. \(58.44\)
6. L. \(10.00\)
7. M. \(584.4\)