1) What quantity of CuI is present in 276 mL of a 0.10 M solution? (Answer in moles: unit=mol.)?

Question

2) What mass of Cu(NO3)2 is present in 276 mL of a 0.40 M solution? (Enter unit as g.)
3) What mass of KNO3 is needed to add to a 500 mL volumetric flask to produce a solution that is 0.40 M? (Enter unit as g.)
4) What is the molarity of a solution produced by dissolving 26 g of KI in a 1000 mL volumetric flask?(Enter unit as M.)

Answer 1

1) M = mol / L ~> mol = M * L
276 mL = 0.276 L
0.10 * 0.276 = 0.028 mol (rounded to two sig figs)

2) 0.40 * 0.276 = 0.1104 mol
Molar mass Cu(NO3)2 = 187.5 g/mol
g = Molar mass * mol = 0.1104 * 187.5 = 21g (rounded to two sig figs)

3) 0.500 * 0.40 = 0.20 mols needed
molar mass KNO3 = 101 g/mol
g = 0.20 * 101 = 20. grams (rounded to two sig figs)

4)molar mass KI = 166g/mol
mol = 26 g / 166 g/mol = 0.16 mol
0.16 mol / 1L = 0.16 M

Answer 2

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