Answer: the scientist should ensure that the volume is measured in dm3(or L), the pressure in atm, the temperature in Kelvin and use the gas constant 0.082atm.dm3.K^-1.mol^-1
X will be in group 5, since if you exchange the valencies of Na with any element on group 5, you will get Na3X
Answer:
3.1°C
Explanation:
Using freezing point depression expression:
ΔT = Kf×m×i
<em>Where ΔT is change in freezing point, Kf is freezing point depression constant (5.12°c×m⁻¹), m is molality of the solution and i is Van't Hoff factor constant (1 For I₂ because doesn't dissociate in benzene).</em>
Molality of 9.04g I₂ (Molar mass: 253.8g/mol) in 75.5g of benzene (0.0755kg) is:
9.04g ₓ (1mol / 253.8g) = 0.0356mol I₂ / 0.0755kg = 0.472m
Replacing in freezing point depression formula:
ΔT = 5.12°cm⁻¹×0.472m×1
ΔT = 2.4°C
As freezing point of benzene is 5.5°C, the new freezing point of the solution is:
5.5°C - 2.4°C =
<h3>3.1°C</h3>
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The answer & explanation for this question is given in the attachment below.
