Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?

1 answer:

5 0

The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).

ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
 ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 °C
 ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 °C
 ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 °C
 ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 °C

Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.

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The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

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= 1.33 g Zn