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How many grams of solute are present in 635mL of 0.450 M KBr?

Sagot :

Eduard22slyidn

Answer:

34.03 g

Explanation:

We'll begin by converting 635 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

635 mL = 635 mL × 1 L / 1000 mL

635 mL = 0.635 L

Thus, 635 mL is equivalent to 0.635 L

Next, we shall determine the number of mole of the solute (KBr) in the solution. This can be obtained as follow:

Volume = 0.635 L

Molarity = 0.450 M

Mole of KBr =?

Mole = Molarity × Volume

Mole of KBr = 0.450 × 0.635

Mole of KBr = 0.286 mole

Finally, we shall determine the mass of 0.286 mole of KBr. This can be obtained as follow:

Mole of KBr = 0.286 mole

Molar mass of KBr = 39 + 80

= 119 g/mol

Mass of KBr =?

Mass = mole × molar mass

Mass of KBr = 0.286 × 119

Mass of KBr = 34.03 g

Thus, the mass of the solute (KBr) in the solution is 34.03 g

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