Question

Felicia dissolved 4.00 g of Co(NO3)2 in enough water to make 100 mL of stock solution. She took 4.00 mL of the stock solution and then diluted it with water to give 275 mL of a final solution. How many grams of NO3- ion are there in the final solution?

Answer 1

calcualte moles of Ca(NO3)2 first

mol = mass/MW

MW = 164.088 g /mol

then

mol = mass/MW = 4/164.088 = 0.02438 mol of Ca(NO3)2

M = mol/L

M =0.02438 /(100/1000) = 0.2438 M of Ca(NO3)2

then she takes

V = 4 ml of stock solution

concnetration is still the same

she dilutes to Vf = 275 ml = 0.275 L

that is

M1V1 = M2V2

0.2438*(4) = M2*(275)

M2 = (0.2438*(4))/(275) = 0.003546 M

this is mol per liter

then

0.003546 in 275 ml

mol = MV = 0.003546*0.275 = 0.00097515 mol of Ca(NO3)2

then

since we have 2 mol of NO3 per molecules of Ca(NO3)2

0.00097515*2 = 0.0019503 mol of NO3- ions

MW of NO3- ion = 62.0049 g/mol

mass = mol*MW = 0.0019503*62.0049 = 0.1209281 g of NO3- present in that amount