How many grams of solute are present in 705 mL of 0.900 M #KB#r?

Question

Caleb Adams · Accepted Answer

Approx. #75*g#....

By definition....#"concentration"-="moles of solute"/"volume of solution"# And so to get the #"moles of solute"#, we take the product.... #"moles of solute"="concentration"xx"volume of solution"#.. Thus, we have the following. #n_"KBr"=0.900*mol*L^-1xx705*mLxx10^-3*L*mL^-1# #=0.6345*mol#... And thus a MASS of #0.6345*cancel(mol)xx119.0*g*cancel(mol^-1)=??*g#

Olivia Anton · Answer

undefined To find the grams of solute, you need to use the formula:

$$\text{grams of solute} = \text{moles of solute} \times \text{molar mass of solute}$$

First, calculate the moles of solute:

$$ \text{moles of solute} = \text{Molarity} \times \text{Volume (in liters)}$$

Then, convert milliliters to liters:

$$705 \text{ mL} = 0.705 \text{ L}$$

Now, use the given molarity (0.900 M) and volume (0.705 L) to find the moles of solute:

$$ \text{moles of solute} = 0.900 \text{ M} \times 0.705 \text{ L} = 0.635 \text{ moles}$$

Next, find the molar mass of KBr:

$$KBr: \text{ K} = 39.10 \text{ g/mol}, \text{ Br} = 79.90 \text{ g/mol}$$

$$Molar \, mass \, of \, KBr = 39.10 + 79.90 = 119.00 \, g/mol$$

Finally, multiply the moles of solute by the molar mass of KBr to find the grams of solute:

$$ \text{grams of solute} = 0.635 \text{ moles} \times 119.00 \text{ g/mol} = 75.65 \text{ grams}$$

So, there are 75.65 grams of solute present in 705 mL of 0.900 M KBr solution.