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What is the mass in grams of KBr in [tex]0.400 \, L[/tex] of a [tex]0.350 \, M[/tex] solution?

Sagot :

Let's solve this problem step-by-step:

1. Identify the given values:

- Volume of the solution ([tex]\( V \)[/tex]) = 0.400 liters (L)
- Molarity of the solution ([tex]\( M \)[/tex]) = 0.350 moles per liter (M)

2. Determine the molar mass of KBr:

KBr is composed of potassium (K) and bromine (Br).

- Atomic mass of potassium (K) = 39.1 g/mol
- Atomic mass of bromine (Br) = 79.9 g/mol

Therefore, the molar mass of KBr ([tex]\( M_{\text{KBr}} \)[/tex]) is:
[tex]\[ M_{\text{KBr}} = 39.1 \, \text{g/mol} + 79.9 \, \text{g/mol} = 119.0 \, \text{g/mol} \][/tex]

3. Calculate the number of moles of KBr in the solution:

Molarity is defined as the number of moles of solute per liter of solution ([tex]\( M = \frac{\text{moles}}{\text{liters}} \)[/tex]).

Rearranging this formula to find the number of moles ([tex]\( n \)[/tex]):
[tex]\[ n = M \times V \][/tex]

Substituting the given values:
[tex]\[ n = 0.350 \, \text{M} \times 0.400 \, \text{L} = 0.140 \, \text{moles} \][/tex]

4. Calculate the mass of KBr in the solution:

The mass of KBr can be determined using the formula:
[tex]\[ \text{mass} = n \times M_{\text{KBr}} \][/tex]

Substituting the values of [tex]\( n \)[/tex] and [tex]\( M_{\text{KBr}} \)[/tex]:
[tex]\[ \text{mass} = 0.140 \, \text{moles} \times 119.0 \, \text{g/mol} = 16.66 \, \text{grams} \][/tex]

Therefore, the mass of KBr in 0.400 liters of a 0.350 M solution is 16.66 grams.