Question

how many grams of mg3(po4)2 are needed to make 3.000 l of a 2.500 m solution? the gfw for magnesium phosphate is 262.87 g.
_1_ mol
( _2_ l)(----------------) = _3_ mol
_4_ l

_5_ g
( _6_ mol)(-------------) = _7_ g
_8_ mol
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 x 10^3 v. 7.500 w. 0.6667
x. 0.200 y. 2.67 z. 2.67 x 10^3 aa. 10^3 bb. 15.31 cc. 30.62
dd. 1971

Explanation:

Step1: Calculate moles using molarity formula

Molarity \(M=\frac{n}{V}\), so \(n = M\times V\). Here \(M = 2.500\space mol/L\) (sub - question 1: B), \(V=3.000\space L\) (sub - question 2: A), \(n=(3.000\space L)\times(2.500\space mol/L)=7.500\space mol\) (sub - question 3: V). The denominator in the formula \(\frac{n}{V}\) is \(1.000\space L\) (sub - question 4: D).

Step2: Calculate mass using molar mass

Mass \(m=n\times M_m\). Here \(n = 7.500\space mol\) (sub - question 6: V), \(M_m=262.87\space g/mol\) (sub - question 5: S), \(m=(7.500\space mol)\times(262.87\space g/mol)=1971.525\approx1971\space g\) (sub - question 7: DD). The denominator in the formula \(m = n\times M_m\) (where \(M_m=\frac{m}{n}\)) is \(1\space mol\) (sub - question 8: D).

Answer:

1. B. 2.500
2. A. 3.000
3. V. 7.500
4. D. 1.000
5. S. 262.87
6. V. 7.500
7. DD. 1971
8. D. 1.000