Constant:
Test tubes
Independent:
<span>volume of gas
Dependent:
</span>
<span>amount of H2O2 </span>
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
Learn more about The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
Answer:
Following are the complete balance of the given equation:
Explanation:
Given equation:
After Balancing the equation:
In the above equation, when the 3 mol Copper sulfate reacts with 2 mol lithium phosphate
, it will produce 2 mol Copper phosphate and 3 mol Lithium sulfate
.
