Question

what is the concentration after 3.000 ml of a 3.000 m nacl solution is diluted with 47.000 ml of water?
identify the variables ( c_d =? ) ( c_c = 3.000 m )
( v_d = 1 ml ) ( v_c = 3.000 ml )
first put in the symbols, then put in the numbers
( c_d=\frac{c_c 2}{4}=\frac{(3.000 m)(3 ml)}{5 ml}=0.1800 m )

1. what is the dilution factor? first put in the symbols, then put in the numbers

dilution ( =\frac{v_c}{8}=\frac{6 ml}{9 ml}=\frac{7}{50} )
a. 3.000 b. 2.500 c. 10.00 d. 1 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\times10^3 ) v. 7.500 w. 0.6667 x. 0.200 y. 2.67
z. ( 2.67\times10^3 ) aa. ( 10^3 ) bb. ( 10^{-3} ) cc. ( c_c ) dd. ( c_d )
ee. ( v_c ) ff. ( v_d )

Explanation:

Step1: Calculate the final volume ($V_D$)

The final volume is the sum of the initial solution volume and the water volume.
$V_D=3.000 + 47.000=50.00$ mL

Step2: Use the dilution formula $C_CV_C = C_DV_D$ to find $C_D$

We know that $C_D=\frac{C_CV_C}{V_D}$. Substituting $C_C = 3.000$ M, $V_C=3.000$ mL and $V_D = 50.00$ mL.
$C_D=\frac{(3.000\space M)\times(3.000\space mL)}{50.00\space mL}=0.1800$ M

Step3: Calculate the dilution factor

The dilution factor formula is $\text{dilution factor}=\frac{V_D}{V_C}$. Substituting $V_D = 50.00$ mL and $V_C=3.000$ mL.
$\text{dilution factor}=\frac{50.00\space mL}{3.000\space mL}\approx16.67$ (but from the given options and the formula structure in the problem: using $\text{dilution factor}=\frac{V_D}{V_C}$, with $V_D = 50.00$ (L) and $V_C=3.000$ (A))

Answer:

1. L (50.00)
2. CC ($C_C$)
3. A (3.000)
4. FF ($V_D$)
5. L (50.00)
6. A (3.000)
7. 16.67 (but from options and formula structure: using $\frac{V_D}{V_C}$, so with $V_D = 50.00$ (L) and $V_C=3.000$ (A))
8. FF ($V_D$)
9. L (50.00)