Question

how many grams of mg₃(po₄)₂ are in 10.00 ml of a 2.500 m solution?
the gfw for magnesium phosphate is 262.87 g.
\\( (10.00 \\, \text{ml})(\frac{\\_\\_1\\_\\_ \\, \text{l}}{\\_\\_3\\_\\_ \\, \text{ml}})=\\_\\_2\\_\\_ \\, \text{l} \\)
\\( (\\_\\_6\\_\\_ \\, \text{l})(\frac{\\_\\_4\\_\\_ \\, \text{mol}}{\\_\\_8\\_\\_ \\, \text{l}})=\\_\\_7\\_\\_ \\, \text{mol} \\) \\( (\frac{\\_\\_5\\_\\_ \\, \text{g}}{\\_\\_9\\_\\_ \\, \text{mol}})=6.572 \\, \text{g} \\)
a. 3.000
b. 2.500
c. 10.00
d. 1.000
e. 2.000
f. 9.000
g. 6.000
h. 8.000
i. 0.02500
j. 47.00
k. 5.00
l. 50.00
m. 25.0
n. 0.02041
o. 97.98
p. 30.62
q. 6.572
r. 0.6124
s. 262.87
t. 0.1800
u. 3.000×10³
v. 7.500
w. 0.6667
x. 0.200
y. 2.67
z. 2.67×10³
aa. 10³
bb. 10⁻⁶
cc. 30.62
dd. 1971
ee. 0.01000

Explanation:

Step1: Convert volume from mL to L

We know that \(1\space L = 1000\space mL\). So, for the conversion factor, \(\frac{1\space L}{1000\space mL}\).
\((10.00\space mL)\times(\frac{1\space L}{1000\space mL})=0.01000\space L\)

Step2: Calculate the number of moles

Using the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, and \(V\) is volume). The concentration \(C = 2.500\space M=\frac{2.500\space mol}{1\space L}\), and \(V = 0.01000\space L\).
\((0.01000\space L)\times(\frac{2.500\space mol}{1\space L}) = 0.02500\space mol\)

Step3: Calculate the mass

Using the formula \(m=n\times M\) (where \(m\) is mass, \(n\) is moles, and \(M\) is molar mass). The molar mass \(M = 262.87\space g/mol\), and \(n=0.02500\space mol\).
\((0.02500\space mol)\times(\frac{262.87\space g}{1\space mol})=6.572\space g\)

Answer:

1. D. \(1.000\)
2. EE. \(0.01000\)
3. AA. \(10^{3}\)
4. B. \(2.500\)
5. S. \(262.87\)
6. EE. \(0.01000\)
7. I. \(0.02500\)
8. D. \(1.000\)
9. I. \(0.02500\)