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

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

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.