Question

question 9 (1 point)
what is the concentration after a 1:25 dilution of a 5.00 m nacl solution?
first put in the symbols, then put in the numbers
vc 1
cd=(5.00m)(---------)=(5.00m)(---------) = 2 m
3 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. 3.000×10³ v. 7.500 w. 0.6667 x. 0.200 y. 2.67
z. 2.67×10³ aa. 10³ bb. 10⁻³ cc. cc dd. cd
ee. vc ff. vd gg. 4.00

Explanation:

Step1: Dilution formula

The dilution formula is \(C_D = C_C\times\frac{V_C}{V_D}\). For a \(1:25\) dilution, \(V_D = 25V_C\), so \(\frac{V_C}{V_D}=\frac{V_C}{25V_C}=\frac{1}{25}\).

Step2: Substitute values

Given \(C_C = 5.00\space M\), then \(C_D=(5.00\space M)\times\frac{1}{25}\).

Step3: Calculate

\(C_D=\frac{5.00}{25}= 0.200\space M\)

Answer:

1. \(1\)
2. \(0.200\)
3. \(25.0\) (corresponds to \(V_D\) in the formula \(\frac{V_C}{V_D}\))
4. \(X\) (since \(C_D = 0.200\space M\) which is option \(X\))