Determine the molality of an aqueous solution that is 15.5 percent urea by mass.

1 answer:

3 0

15.5% by mass is equivalent 15.5 g urea in 100 g solution or 155 g urea in 1 kg solution. 

we know that molality = moles solute / kg solvent

moles solute = 155 g / 60 g/mol = 2.58 moles urea

Since there are 155 g urea in 1000g solution, hence the solvent is 845 g or 0.845 kg

So:
molality = 2.58 / 0.845 = 3.06 m

You might be interested in

Name two human actions needed to slow climate change

NeTakaya

Burning less fossil fuels and riding bikes or walking instead of cars.

Explanation:

7 0

3 years ago

Read 2 more answers

When is Roblo..x coming back on? :////

Kryger [21]

I've no idea about it...

4 0

2 years ago

Read 2 more answers

(a) a 0.2 m potassium hydroxide solution is titrated with a 0.1 m nitric acid solution. (i) balanced equation: (ii)what would be

Stells [14]

$\text{KOH} (aq) + \text{HNO}_3 (aq) \to \text{KNO}_3 (aq) + \text{H}_2\text{O} (l)$

The solution shall contain only $\text{KNO}_3 (aq)$ (and water) at the equivalence point. Both potassium hydroxide and nitric acid exist as strong electrolytes. As a result, $\text{KNO}_3 (aq)$ , the salt derived from a reaction between the two species would undergo hydrolysis of a negligible extent. This neutralization reaction therefore be neutral at the equilibrium point.

The question states that the solution is "titrated with a ... nitric acid solution" indicating that $\text{HNO}_3$ is added to the initially-basic solution. PH value of the solution would keep decreasing as the volume of the acid added increases. The final solution would be acidic as it contains not only water and $\text{KNO}_3 (aq)$ , but some $\text{HNO}_3$ as well. Bromothymol blue would therefore demonstrates a yellow color, the color it present in an acidic solution, at the end of the titration.