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
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Answer:
most likely c
Explanation:
the volume is increased by the quickly moving water particles
Given
V1 = 35L
T1 = 20.00 C = 293.0 K
V2 = ???
T2 = 45 C = 318 C
Formula
V1/T1 = V2/T2
Sub and Solve
45/293 = x / 318 Cross multiply
45*318 = 293x Divide by 293
45*318/293 = x
14310 / 293 = x
x = 48.84
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Answer:
55.95K
Explanation:
its obtained from Q=MCT where
Q-amount of heat
M-mass of water
C-specefic heat capacity of water
T-change in temperature
