Determine the number of grams of Potassium Iodate present in 10.0 g of a 0.267 m Potassium Iodate solution.

Is the m or M?

it says "m" but I am not sure that is right.

OK, we'll assume that is m and 0.267 molal KIO3.

0.267m means 0.267 mols KIO3/1000 g solvent.
0.267 mols KIO3 x molar mass KIO3 = grams KIO3 = approx 57 g but you need to that more accurately as well as the calculations that follow.
Then total solution is g KIO3+ 1000g solvent = 1000 + about 57 = about 1057g so in essence we have
57g KIO3/1057 g solution. You want 10 g of that so g KIO3 in 10 g is
about 57 x (10/1057) = ? g KIO3 in 10 g of 0.267 m KIO3.

To determine the number of grams of Potassium Iodate present in the given solution, we need to use the equation:

\( \text{{molarity (M)}} = \frac{{\text{{moles of solute}}}}{{\text{{liters of solution}}}} \)

First, let's calculate the number of moles of Potassium Iodate in the solution:

\( \text{{Molarity (M)}} = \frac{{\text{{moles of solute}}}}{{\text{{liters of solution}}}} \)

Rearranging the equation, we have:

\( \text{{moles of solute}} = \text{{Molarity (M)}} \times \text{{liters of solution}} \)

Given that the molarity of the solution is 0.267 M, we can substitute this into the equation.

However, we need to convert grams of the solution to liters using the density of the solution. Once we have the liters, we can substitute this value into the equation to calculate the moles of solute.

Therefore, we need the density of the solution in order to calculate the volume in liters.

Do you have the density of the solution?