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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.

• chem -

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.

• chem -

a was incorrect

• chem -

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.

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