Calculate the number of moles of solute present in each of the following solutions.

a) 39.0 mg of an aqueous solution that is 1.00 m NaCl

b)90.0 g of an aqueous solution that is 2.10% sucrose (C12H22O11) by mass.

a.

1.00 m NaCl means 1.00 mol NaCl/kg solvent (1000 g solvent)
1 mol NaCl = about 58g but you need to do it more accurately than that.
The solution then is 1000 g solvent + 58 g solute = 1058 g total. You want 39.0 mg of this solution.
1 mol NaCl x (0.039g/1058g)= ?mol NaCl.

b. Is that by mass or some other %. It makes a difference. I'll assume by mass.
2.10% sucrose means 2.10g sucrose/100 g soln.
g sucrose in 90g of soln = 2.10 x 90/100 = ?
Convert that to mols.

a was incorrect

The method is correct. I expect you didn't correct the math part. I reminded you that you needed to do the math and do it more accurately than in my example. If you will post your work I will find your error.

To calculate the number of moles of solute present in a solution, you need to use the formula:

moles of solute = mass of solute / molar mass of solute

In both cases, we need to convert the mass of the solute to grams if it is given in a different unit.

a) Let's calculate the number of moles of NaCl in 39.0 mg of the solution.

First, we need to convert the mass of NaCl to grams by dividing by 1000:
39.0 mg = 39.0 mg / 1000 = 0.039 g

The molar mass of NaCl is 58.44 g/mol (the atomic mass of sodium is 22.99 g/mol, and chlorine is 35.45 g/mol).

Now we can use the formula to calculate the number of moles of NaCl:
moles of NaCl = 0.039 g / 58.44 g/mol = 0.000667 mol (rounded to 4 decimal places)

Therefore, there are approximately 0.000667 moles of NaCl in the given solution.

b) Let's calculate the number of moles of sucrose (C12H22O11) in 90.0 g of the solution.

The percentage by mass of sucrose can be converted to grams by multiplying it by the mass of the solution:
mass of sucrose = 2.10% * 90.0 g = 1.89 g

The molar mass of sucrose (C12H22O11) is 342.30 g/mol.

Now we can use the formula to calculate the number of moles of sucrose:
moles of sucrose = 1.89 g / 342.30 g/mol = 0.00552 mol (rounded to 5 decimal places)

Therefore, there are approximately 0.00552 moles of sucrose in the given solution.