The top left corner shows the atomic number which is 3.
As long as the chemical is not used up in the reaction the answer is true
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
213034 torr is the osmotic pressure.
Explanation:
osmotic pressure is calculated by the formula:
osmotic pressure= iCrT
where i= no. of solute
c= concentration in mol/litre
R= Universal Gas constant
T = temp
It is given that solution is 3% which is 3gms in 100 ml.
let us calculate the concentration in moles/litre
3gm/100ml*1000ml/1L*1mol NaCl/55.84g NaCl
= 5.372 gm/litre
Putting the values in the formula, Temp in Kelvin 318.5K
osmotic pressure= 2*5.372*0.083 * 318.5 Gas constant 0.083
= 284.023 bar or 213018 torr. c= 5.372 moles/L
i=2 for NaCl
