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:
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Explanation:
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Answer:
grams KI needed = 31.9 grams
Explanation:
? g KI + 500 g water => 6.0% KI solution
let x = grams KI needed.
Percent = part/total x 100% = (x/x+500)·100% = 6.0%
Solve for 'x' ...
x / x + 500 = 6/100 = 0.06 => x = 0.06(x + 500) => x = 0.06x + 30
x - 0.06x = 30 => 0.94x = 30 => x = 30/0.94 = 31.9 gms KI needed.
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