Question

question 10 (1 point)
the concentration of a solution was 0.100 m after a 1:25 dilution. what was the
original concentration?
first put in the symbols, then put in the numbers
_1_
cc = (0.100 m)(-----) = (0.100 m)(-----) = _3_ m
_2_
vc
_4_
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. 0.06124 s. 262.87
t. 0.1800 u. 3.000 x 103 v. 7.500 w. 0.6667
x. 0.200 y. 2.67 z. 2.67 x 103 aa. 103 bb. 10-3
cc. cc dd. cd ee. vc ff. vd gg. 4.00

Explanation:

Step1: Use the dilution formula

The dilution formula is \(C_1V_1 = C_2V_2\). Here, assume the original volume \(V_1 = 1\) (arbitrary unit for ratio - based on 1:25 dilution, if \(V_1\) is 1 part, \(V_2\) is 25 parts). \(C_1\) is the original concentration (unknown), and \(C_2=0.100\space M\)

Step2: Rearrange the formula to solve for \(C_1\)

From \(C_1V_1 = C_2V_2\), we get \(C_1=\frac{C_2V_2}{V_1}\). Substituting \(V_2 = 25\) (since 1:25 dilution means final volume is 25 times the initial volume if initial volume \(V_1 = 1\)), \(C_2 = 0.100\space M\) and \(V_1=1\) into the formula: \(C_1=\frac{0.100\space M\times25}{1}=2.500\space M\)

Answer:

B. 2.500