Question

lesson 5 assignment: concentration ii
part 1

1. a 180 - ml disinfectant solution contains 85 ml of isopropyl alcohol. determine the % v/v concentration of this solution.
2. if 114 g of sucrose is dissolved in water to make a 950 - ml solution, determine the % w/v concentration of the solution.
3. a 0.51 - kg solution contains 87 mg of potassium iodide. calculate the % w/w concentration of this solution.
4. 0.30 kg of magnesium sulfate is dissolved in water to make a 92 - kg solution. express the concentration in ppm.
5. determine the volume of methanol required to prepare 2.4 l of a 38% v/v solution.

Explanation:

Step1: Recall the formula for % V/V

The formula for volume - by - volume percentage concentration is $\%V/V=\frac{V_{solute}}{V_{solution}}\times100\%$.
For a 180 - mL disinfectant solution with 85 mL of isopropyl alcohol, $\%V/V=\frac{85}{180}\times100\%$.
$\%V/V=\frac{8500}{180}\approx47.2\%$

Step2: Recall the formula for % W/V

The formula for weight - by - volume percentage concentration is $\%W/V=\frac{m_{solute}}{V_{solution}}\times100\%$.
If 114 g of sucrose is dissolved in water to make a 950 - mL solution, $\%W/V=\frac{114}{950}\times100\%$.
$\%W/V=\frac{11400}{950} = 12\%$

Step3: Recall the formula for % W/W

First, convert 87 mg to kg. Since 1 kg = 1000000 mg, 87 mg=$87\times10^{- 6}$ kg.
The formula for weight - by - weight percentage concentration is $\%W/W=\frac{m_{solute}}{m_{solution}}\times100\%$.
For a 0.51 - kg solution with 87 mg ($87\times10^{-6}$ kg) of potassium iodide, $\%W/W=\frac{87\times10^{-6}}{0.51}\times100\%$.
$\%W/W=\frac{87\times10^{-4}}{0.51}\approx0.017\%$

Step4: Recall the formula for ppm

The formula for parts - per - million (ppm) is $ppm=\frac{m_{solute}}{m_{solution}}\times10^{6}$.
0.30 kg of magnesium sulfate is dissolved in water to make a 92 - kg solution.
$ppm=\frac{0.30}{92}\times10^{6}=\frac{300000}{92}\approx3260.9$ ppm

Step5: Recall the formula for volume of solute

The formula for $\%V/V=\frac{V_{solute}}{V_{solution}}\times100\%$. Rearranging for $V_{solute}$, we get $V_{solute}=\frac{\%V/V}{100\%}\times V_{solution}$.
To prepare 2.4 L of a 38% V/V solution of methanol, $V_{solute}=\frac{38}{100}\times2.4$ L.
$V_{solute}=0.38\times2.4 = 0.912$ L or 912 mL

Answer:

1. Approximately 47.2%
2. 12%
3. Approximately 0.017%
4. Approximately 3260.9 ppm
5. 912 mL