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 mathrm{m} ) ( v_{d}= ) _1_ ml ( v_{c}=3.000 mathrm{~ml} ) first put in the symbols, then put in the numbers ( c_{c} ) _2_ ( (3.000 mathrm{m})( ) _3_ ml) ( c_{d}=\frac{}{} ) _4_ ( =\frac{}{} ) _5_ ml ( =0.1800 mathrm{m} ) 2. what is the dilution factor? first put in the symbols, then put in the numbers ( \begin{array}{ll}v_{c} & \text { _6_ } mathrm{ml} \\ \text { dilution }=\frac{}{} \text { _8_ } & \text { _9_ } mathrm{ml}end{array}=\frac{}{} ) _7_ ( =\frac{}{} ) 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 \times 10^{3} ) v. 7.500 w. 0.6667 x. 0.200 y. 2.67 z. ( 2.67 \times 10^{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 \(V_D\) is the sum of the initial volume \(V_C\) and the volume of water added. So \(V_D=V_C + \text{volume of water}=3.000\space mL+ 47.000\space mL = 50.000\space mL\)

Step2: Use the dilution formula \(C_CV_C = C_DV_D\)

We know \(C_C = 3.000\space M\), \(V_C=3.000\space mL\), \(V_D = 50.000\space mL\). Rearranging the formula for \(C_D\) gives \(C_D=\frac{C_CV_C}{V_D}\)
Substitute the values: \(C_D=\frac{3.000\space M\times3.000\space mL}{50.000\space mL}\)

Step3: Calculate the dilution factor

The dilution factor formula is \(\text{Dilution factor}=\frac{V_D}{V_C}\)
Substitute \(V_D = 50.000\space mL\) and \(V_C=3.000\space mL\)

Answer:

1. \(V_D = 50.00\space mL\) (L), \(C_CV_C\) (\(CC\times EE\)), \(V_C = 3.000\space mL\) (A), \(V_D\) (FF), \(V_D=50.00\space mL\) (L)
2. \(V_C = 3.000\space mL\) (A), \(V_D = 50.00\space mL\) (L), \(\text{Dilution factor}=\frac{V_D}{V_C}=\frac{50.00\space mL}{3.000\space mL}\approx16.67\) (but if we consider the formula structure in the problem, for the blanks: \(V_C\) (EE), \(V_C = 3.000\space mL\) (A), \(V_D\) (FF), \(V_D = 50.00\space mL\) (L))