A student dissolves 18.9 grams of Zn(NO3)2 [molar mass = 189] in enough water to make 500.0 of solution. A 100.0 ml portion of this solution is diluted to 1000.0 ml.What is the [NO3¯ ] in the second solution?
A) 0.0100M B) 0.0200M C) 0.0300M
D) 0.0400M E) 0.0500M
Convert 18.9 g Zn(NO3)2 to moles. moles = grams/molar mass
Nitrate ion will be twice that since there are two moles NO3^- to each mole of Zn(NO3)2.
Placing that in 500 mL will be
M = moles/0.500 L = ?? M.
You then dilute 100 mL of that soln to 1000 mL; therefore, the new concn will be ??M x (100/1000) = zz M.
To find the [NO3¯] in the second solution, we need to calculate the number of moles of NO3¯ present in the 100.0 ml portion and then divide by the volume of the final solution.
First, we need to find the number of moles of Zn(NO3)2 in the original solution. We can use the equation:
moles = mass / molar mass
molar mass of Zn(NO3)2 = 189g/mol
mass of Zn(NO3)2 in solution = 18.9g
moles of Zn(NO3)2 = 18.9g / 189g/mol = 0.1 mol
Next, we need to find the number of moles of NO3¯ in 0.1 mol of Zn(NO3)2. In the compound Zn(NO3)2, there are two NO3¯ ions for every one molecule of Zn(NO3)2.
moles of NO3¯ = 2 moles Zn(NO3)2 * (1 mole NO3¯ / 1 mole Zn(NO3)2) = 0.2 moles
Now, we can calculate the concentration of NO3¯ in the 100.0 ml portion of the solution. Remember, concentration is moles/volume.
concentration of NO3¯ = moles of NO3¯ / volume
volume of the 100.0 ml portion = 100 ml = 0.1 L
concentration of NO3¯ = 0.2 moles / 0.1 L = 2 M
Finally, we need to find the concentration of NO3¯ in the final solution by dividing the concentration in the 100.0 ml portion by the dilution factor.
dilution factor = final volume / initial volume = 1000 ml / 100 ml = 10
concentration of NO3¯ in final solution = 2 M / 10 = 0.2 M
Therefore, the [NO3¯] in the second solution is 0.2 M. So, the correct answer is not listed in the choices given.