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
To determine the molarity of [NO3¯] in the second solution, we need to use the concept of dilution.
First, let's calculate the number of moles of Zn(NO3)2 in the first solution:
Mass of Zn(NO3)2 = 18.9 g
Molar mass of Zn(NO3)2 = 189 g/mol
Number of moles of Zn(NO3)2 = mass / molar mass = 18.9 g / 189 g/mol = 0.1 mol
Next, let's calculate the volume of the 100.0 ml portion of the first solution that is diluted:
Volume of 100.0 ml portion = 100.0 ml
Now, we need to find the molarity of the 100.0 ml portion of the first solution:
Molarity of the 100.0 ml portion = moles / volume (in liters)
Since the volume is given in milliliters, we need to convert it to liters.
Volume of 100.0 ml portion in liters = 100.0 ml / 1000 ml/L = 0.1 L
Molarity of the 100.0 ml portion = 0.1 mol / 0.1 L = 1.0 M
Finally, we dilute the 100.0 ml portion to 1000.0 ml, which is a 10-fold dilution.
This means that the molarity of the diluted solution is 1/10th of the original molarity.
Molarity of the diluted solution = 1.0 M / 10 = 0.1 M
The molarity of [NO3¯] in the second solution is 0.1 M.
Therefore, the correct answer is A) 0.0100 M.